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Transcript
91 PROJECT Without Wax PROJECT 1.1 Aim : To find the difference of frequency of the two given turning forks by method of beats & to find out the tuning fork of higher frequency. S. No. of No. beats x1 Time taken No. of beats/ second y1 = Apparatus : Two tuning forks A and B of nearly equal frequency. These forks should be attached on different resonance boxes, wax, stop watch, rubber pad etc. 1. 10 2. 10 3. - Principle : If two sound waves of equal amplitude & nearly equal frequency moving in same direction superimpose then their resultant intensity gradually increases & decreases. This phenomenon is called formation of beats. One maximum & one minimum intensity of sound constitute 1 beat. Beat frequency is equal to difference in frequency of both the tuning forks. Let frequency of tuning fork A be n 1 & that of B be n2. Where (n1 „ n2) Then no. of beats per second = n1 – n2 (if n1 > n2) or n2 – n1 (if n2 > n1). Therefore we can find the difference in their frequencies by sounding both of them together. But which is greater out of n1 & n2? It can be found by loading one of them with wax. If a prong of a tuning fork is waxed then its frequency decreases. Average No. of beats/second = y 1 = n1 n2 second y1 = Therefore if no. of beats per second increases on waxing A then n 1 > n2 but if no. of beats per second decreases on loading then n1 < n2. Average No. of beats/second = y 1 = n1 n2 t (sec.) x1 t Mean y1 = = Difference in frequencies of the two tuning forks. = - - Hz. After Waxing S. No. of Time taken No. beats x1 t (sec.) No. of beats/ x1 t Mean y2 = = Difference in frequency of the two tuning forks = ................. Hz Method : (i) Place both the resonance boxes with tuning forks close to each other. (ii) Strike both of them on rubber pad one after the other gently and hear the beats. (iii) Note down time for 10 beats with the help of stop clock. (iv) Repeat above steps (ii) & (iii) 4 times. Now on turning fork A put some wax so that its frequency decreases. v) Now repeat (ii) & (iii) i.e. again find time for 10 beats. Observations : Least count of stop clock = .... Second. Tuning fork with sound box Result : (i) Difference in frequencies of the two tuning forks = ... Hz (ii) As on waxing difference in frequencies has .......... therefore frequency of A is ............ than frequency of B. 92 Precautions : i) The difference in the frequencies of the two tuning forks should not be more than 10. ii) Experiment should be done in a calm environment, so that beats are heard clearly. iii) Both tuning forks should be struck with nearly the same force & should be sounded together. iv) Both tuning forks should vibrate for about 10-12 seconds. v) Stop clock should be started either at minimum or maximum intensity of sound. vi) Both tuning forks or their resonance boxes should be kept as close as possible so that beats are heard clearly. 5. What are the essential conditions for beat formation? 6. What will happen to the frequency of the tuning fork on waxing? If one prong is filed how will the frequency of the tuning fork change? 7. How many beats per second will be produced when two tuning forks of frequencies 246 and 248 are sounded together? ANSWERS 3. No, interference will be produced. 4. When two waves of exactly same frequency move in the same direction and in the same medium, then due to their superposition, redistribution of energy takes place. This phenomenon is called interference. VIVA-VOCE QUESTIONS 1. What do you understand by beats? 5. 2. When is 1 beat completed? If two tuning forks of frequencies n1 & n2 are used, how many beats will be obtained per second? There should be a small difference (less than 10) in frequencies of the two tuning forks. 6. It will decrease. By filing one arm of the tuning fork, frequency will be increased. 3. If both tuning forks are of the same frequency will the beats be heard? 7. 248 – 246 = 2 beats per second. 4. What is interference? 93 PROJECT 1.2 Aim : To prove that n l = constant with the help of resonance tube, where n is the frequency and l is the length of the resonating air column. 3. Apparatus : A glass tube having about 3 cm internal diameter open at both ends, glass jar in which tube can be dipped, one iron stand, half metre scale. Tuning forks of frequencies 256, 480 and 512 Hz. Principle : When a tuning fork is placed near a closed organ pipe then stationary waves are produced. Antinodes are formed near the open end and nodes are formed at the closed end. Distance between a node and the nearest antinode = l = λ/4 Resonance Tube Let the length of organ pipe be l + then l = λ/4. If frequency of tuning fork= natural frequency of air column then resonance will be produced in the air column. If v is velocity of sound then frequency of resonating air column. n= pad and place it near the open end of the resonance tube. (Fig.) Some sound may be heard. v v v = or l = λ 4l 4n As velocity of sound is constant at a fixed temperature l ∝ 1/n or nl = content Hence length of vibrating column is inversely proportional to frequency of the tuning fork which is equal to that of air column at resonance. Method : 1. Assemble the apparatus as shown in fig. 2. Strike the tuning fork gently on the rubber Increase the length of air column of the resonance tube and repeat step (2). You will see that intensity of sound increases. Slowly pull the tube up till maximum sound is heard. At this stage air column is resonating such that its natural frequency= frequency of tuning fork. Note down this length. First Resonance 4. Repeat steps 2 & 3 with another tuning fork. 5. Repeat steps 2 & 3 with 3rd tuning fork. 6. Record your observations in a table. 7. Find n for each tuning fork. You will find that constant. Note down room temperature. Observations : Make observation tables as in experiment No. 7 Results : It is found that for resonance column nl remains constant. Precautions : Same as in experiment No. 7 Discussion : n l is constant only at a constant temperature. If room temperature changes, value of nl will also change. 94 PROJECT 2 Aim : To determine spring constant of the given spring. Apparatus : Spring, pan, scale & pointer arrangement, 50 gram weights. Description of apparatus Suspend a light spring from a rigid support & on the other end a light pan. Near the pan attach a pointer. Pointer is able to move on the scale (fig. 10.10) Fig. 10.11 : Extension of spring Since g = constant k l ∝m Therefore extension is directly proportional to m. Hence the graph is a straight line. Slope of the line = Fig. 10.10 : Spring scale arrangement ∆l g = S, But S = k ∆m Principle : A light spring of length L is hanging from a rigid support in a vertical position. If a mass m is attached at the lower end, the spring increases in length by l (fig. 10.11). In this position restoring force F acts on the spring. According to Hooke’s Law F = –k l ..... (1) Where k = spring constant. As force F acts opposite to displacement, therefore in eq. (i) negative sign is used. Here F is acting in vertically upward & F = mg in vertically downward direction. As object is in equilibrium. F + mg = 0 or – kl + mg = 0 or kl = mg () l= g m or k \ Spring constant k = g S METHOD i) ii) Arrange the apparatus as shown in fig (10.10) Note down the position of spring by pointer without any weight. iii) After that put 50 gram weights one by one on the pan & every time note down position of the pointer. After placing weights on pointer wait for some time before taking the observations. Now remove the weights one by one and note down position of pointer everytime after waiting for some time. 95 Observations : Least count of these scale = ....... cm S. No. Weights (force) 1. 2. 3. 4. 5. 6. 7. 0 50 g 100 g 150 g 200 g 250 g 300 g Reading of the pointer Loading Unloading Mean Total extension (mm) Extension for 50 gm wt. (mm) -- --- x1–x0 x2–x0 x3–x0 x4–x0 x5–x0 x6–x0 x1–x0 x2–x1 x3–x2 x4–x3 x5–x4 x6–x5 x0 x1 x2 x3 x4 x5 x6 mean l = Calculations : Slope of line = force constant k = PN MN g 980 = dyne/cm S S = ............... Nm–1 P ››› a spring? 2. Write formula and unit of force constant. 3. Find resultant force acting on an object, suspended by a spring in equilibrium position. 4. Which spring has less force constant- light or soft? 5. If a load is pulled and then released, which kind of motion does the load perform? 6. How observations are taken while loading or unloading? 7. Why do we take observations after sometime after loading or unloading? 8. On what factors does the extension in the spring depend? ∆t M l (cm) θ ∆ m N l (cm) → ANSWERS 1. Force constant is numerically equal to the force to produce a unit extension in the spring. Its unit is newton/metre. Precautions : 2. k= i) Spring should be vertical. 3. 0 ii) Do not put too much load on the spring otherwise the extension of the spring will become irregular. 4. Soft spring. 5. Simple harmonic motion. 6. (i) It corrects error due to elastic fatigue, (ii) it tells us whether elastic limit has been crossed or not. If load has crossed elastic limit then observations in loading and unloading will differ widely. Such observations are left out. 7. So that vibrations of the spring stop. 8. (1) On force constant of the spring. l – m graph Result : Force constant of the spring= N/m iii) Suspend the spring from a rigid support. iv) After loading or unloading let the pointer become steady & only after that take the observation. v) Do not load or unloaded the spring with a jerk. VIVA-VOCE QUESTIONS 1. What do you understand by force constant of mg newton/metre. l (2) On load suspended. 96 PROJECT 3 Aim : To make a comparative study of inertial and gravitational masses. Apparatus : Inertia balance, standard weights stop watch, unknown mass, stand, scale of 30 cm, spring with a hook, card board, paper etc. Now place unknown mass (object) on the platfrom. Find its time period, and from calibration curve find corresponding mass. We measure the same mass by spring balance also. Now place several known masses on the spring balance and find the length of spring balance. Plot a calibration curve (m – l graph). Now place unkown mass on spring balance find l for unknown mass. Corresponding to l find m from the calibration curve. From observations it is found that inertial mass = gravitational mass. Method 1. Place a mass of 10 gram on the platform of the inertia balance. 2. Produce horizontal vibrations of the blade and find time for 15 vibrations. 3. Now repeat (1) and (2) for 20, 30, 40 & 50 gram. Note down time and masses in a tabular form. Spring Balance : As shown in the fig., a 30cm. scale is attached to a stand and a spring is hung as shown. Spring should have a hook. Mass of hook should be known and the object should be suspended from the hook. A cardboard pointer P is attached to the spring with the help of which scale could be read. 4. Find time period for each mass. 5. Keep unknown mass on platform and find t for 15 vibrations. These observations are for intertial mass. 6. Now place 10, 20, 30, 40 & 50 gram weights on spring balance and find length of spring from scale. In the end hang the same unknown mass from the spring balance and note down the increase in length. Find corresponding mass fromthe calibration curve. Principle : Time period of an object performing S.H.M. is Observation : Least cound of stop watch = second. Inertia Balance Description of apparatus : Inertia balance - Fix a steel strip (hacksaw blade) as shown in fig. Attach a platform on free end of strip. Platform can be made from cardboard or thick paper. Object is kept on it. On pressing, the strip should atleast perform 15 vibrations (fig.) T =2π m m or T2 = 4 π2 or T2 k k ∝ m (Since 4π2/k = constant) Where m = inertial mass and k=spring constant. If a graph is plotted between T2 and m, it will be a straight line. First we take few known masses and find their time periods and plot a calibration curve. 1. For inertia balance : S. No. 1. 2. 3. 4. 5. 6. Mass on plaftorm m gram Time for 15 vib. t Sec. Time period T2 = T=t/15 (second)2 second Unknown m1 t’ T’ 97 2. For spring balance : S. No. 1. 2. 3. 4. 5. 6. Total Mass on hanger m (gram) Reading of scale l (cm) Unknown m1 l1 Calculations : i) For inertial mass : Draw T 2–m graph where m is known mass. Draw graph in the form of best fit straight line. It is the calibration curve for this balance. From it, mass corresponding to (T2)2 can be read. That is Interital mass = ....... gravitational mass Precautions : 1. Hacksaw blades should be flexible so that they can make a larger number of vibrations before coming to rest. 2. Least count of stop watch as the measurements. Discussion : Already calibrated spring bvalance is available in the laboratory. Spring balance used in this activity should be checked by the laboratory spring balance. In fig. value of T for m = 0 shown time period for an empty platform and in fig value of l for m = 0 indicates zero of scale, for spring, without load. From slope of fig. rigidity of steel strip and from slope of fig. force constant of spring can be calculated. VIVA-VOCE QUESTIONS 1. What is meant by inertial mass? 2. What is gravitational mass? 3. ‘Both these masses are equivalent’ What is this statement called? 4. What is the importance of this principle of equivalence. the value of unknown mass m’ fig. 5. What is the relation between mass & weight? ii) For gravitational mass : Plot a graph between m and l. It is calibration curve for spring balance. Now 6. Which mass is found by physical balance? T2 – m Graph ANSWERS 1. Intertial mass is based on property of inertia of an object. 2. Gravitational mass is based on force of gravity on an object. 3. It is known as principle of equivalence. 4. On the basis of this principle Einstein included gravitation in theory of relativity. 5. Weight = W = mg. 6. Gravitational mass. l – m graph find unknown mass corresponding to l. Result : Inertial mass of object = ........ gram. Gravitational mass of object = .......... gram. 98 PROJECT WORK 4 A DISPLACEMENT, VELOCITY AND TICKER TIMER 1 Introduction In many experiments you will require the knowledge of velocity, displacement, distance and other related physical quantities. Therefore, first we will describe these physical quantities in brief. object is ∆s then its instantaneous speed. U = Lim ∆t → 0 ∆s ds = ∆t dt Its S.I. units is ms–1, and its dimensions are (MºLT ) –1 You will have to use a device named ticker timer in 3 experiments and 1 activity, therefore, we will discuss, the contruction and principle of this device also. 2 A Brief Discussion of certain kinematical quantities. Distance : Actual path length covered by a moving object in a definite time interval is known as distance traveled by that object. Its S.I. units is metre and its dimensions are MºLTº. It is a scalar quantity. Displacement : Distance traveled by an object between its initial and final positions in a given direction is known as its displacement. It is a vector quantity. Its S.I. units is metre and dimensions are (MºLTº). Velocity : The rate of change of displacement of a moving object is called its velocity. It is also defined as distance traveled by an object in a unit time along a given direction. It is a vector quantity. Its S.I. unit is m–1, and its dimensions are (MºLT –1) Uniform velocity : If an object covers equal displacements in equal intervals of time, it is said to possess a uniform velocity. Instantaneous velocity : Velocity of an object at a given time is known as its instantaneous velocity. If displacement → S of an object changes with time, then instantaneous velocity at time t is given by → V= If positions of a moving object at times t1 and t2 are respectively x(t 1) and x (t2) then displacement in time internal (t2 – t1) is S = x (t2) – x (t1). → dS dt In this time interval distance travelled by the particle can be greater than or equal to displacement. Distance is always positive but displacement may be positive, negative or zero. Uniform motion : If an object covers equal distances in equal intervals of time in a given direction then it is said to possess uniform motion. Distance time graph : For an object having uniform motion the distance time graph is a straight line inclined to time axis. Slope of this line gives the value of the uniform spead of the object. Speed : It is distance traveled by a moving object in a unit time. If an object covers distance’S in time t then its speed = S/t. It is a scalar quantity. Instantaneous speed : Speed of an object at a given time is known as its instantaneous speed. If in a very small interval of time. ∆t, distance traveled by the Velocity Time Graph: (i) In uniform motion velocity time graph is a straight line parallel to time axis (fig) (ii) Area enclosed by the velocity time graph and time axis between t1 and t2 (shaded portion) gives displacement of the object in that time interval, fig. (iii) The position of an object moving with uniform 99 Acceleration : Rate of change of velocity of an object with time is known as its acceleration. It is indicated by a. If value of a is positive then it is called acceleration and if value of a is negative then it is called retardation. In other words, rate of increase of velocity with time is known as acceleration and rate of decrease of velocity with time is known as retardation. –2 Acceleration is vector quantity. Its S.I. unit is ms . Its dimensional formula is [MoLT –2]. If velocity → V is a S can also be written as S = x (t) – x(o) as it is a function of time. It is shown in fig. The slope of the tangent drawn to the curve at any point gives instantaneous velocity of the object at that time. If the object is moving with uniform velocity, its acceleration is zero and the displacementtime graph is a straight line parallel to axis of time, (fig.) Displacement x(t) → velocity at time t is given by x(t) = x (o) + ut, where x (o) is initial position of object and u is uniform velocity of the object. The position time graph is a straight line inclined to the time axis. Slope = Vol. θO Time (t) → function of time then dS is known as instantaneous dt Displacement-time graph (Uniform Velocity) 3 Ticker timer : acceleration. Uniformly accelerated motion : If the velocity of an object changes equally in equal intervals of time then the object is said to have uniform acceleration and motion of the object is called the uniformly accelerated motion. For such a motion velocity- time graph is a straight line inclined to time-axis (fig.) Displacement of the object between time intervals t1 and t2 is equal to the area enclosed by the velocity time graph. It is shown by shaded portion in fig. It is a device to measure short time intervals. With its help positions of objects at an interval of sec. can be recorded on a tape. Construction : Ticker timer has 3 major parts. Ticker, electromagnet, paper tape and carbon paper disk arrangement. Ticker is a steel strip which can vibrate. One end of ticker is a short needle which strikes a small carbon disk when it vibrates. From below carbon disk a paper tape records dots at equal intervals of time. After one complete vibration of the ticker, one dot is marked. The time interval between two successive dots is called a tick. A tick is taken as a unit of time. Time interval between two successive dots is known as time period of vibrations of ticker timer. V – t Uniformly accelerated motion Electro Magnet Displacement time graph of uniformly accelerated motion is a parabola. It is obtained from the relation, Displacement x(t) → 1 S = v(o ) t + at 2 2 ..... (i) Disk of carbon paper Table Hand Displacement - time graph (uniformly accelerated motion) Time (t) 1 50 From step down transformer connected to A.C. mains Ticker Timer Why does the ticker vibrate? The reason is that one end of ticker is fixed with a needle to carbon 100 Current I paper disk and other end is placed between the pole pieces of a U-shaped electromagnet SN. Alternating Current at 6 Volts and 50 Hz is passed through the electromagnet. Therefore ticker starts vibrating sideways. To produce the required A.C. a stepdown transformer is used which has an input of 220 volt and output of 6 volt. Input terminals of step down transformer are connected to A.C. mains. Output terminals of transformer are connected to the electromagnet. Frequency of vibrations of ticker is same as that of : O A B C D Time (t) Graph of Alternating Current A.C. i.e. 50 Hz. Therefore ticker complets 50 vibrations in one second. Therefore time interval between 2 successive dots is 1/50 second. In India the home supply is 220 volt. A.C. at 50 Hz. Since in ticke timer experiments, A.C. & transformer have important applications, we will discuss them in brief. Alternating Current : In A.C. magnitude and direction of current changes periodically with time. In fig. its variation with time is shown. Mathematically current is expressed as to I=Io sin wt, where I is value of current at time t. Io is peak value of current and w is angular frequency. Angular frequency is related to frequency as ω = 2πυ A.C. changes its direction times in 1 sec. In (Fig.) OB & BD are two cycles of A.C. In OA, direction of current is positive and in AB it is negative. In BD also current is positive for one half cycle and negative for the other half cycle. Because current is continuously changing its direction, hence it is known as alternating current (A.C.). Home supply of A.C. is at a frequency of 50 Hz. In 1 second the current completes 50 cycles. Therefore ticker also has a frequency of 50 Hz. Transformer : It is a device to convert A.C. of a given voltage to A.C. of another voltage. Transformers are of two types. (i) Step-down transformer and (ii) Step-up transformer. Step down transformer changes high voltage. A.C. to low voltage A.C. Step up transformer changes low voltage. A.C. to high voltage. A.C. for ticker timer 220 volt. A.C. is converted to 6 volt. A.C. Therefore, we use a step down transformer. 101 PROJECT 4.1 (A) Aim : to demonstrate the motion of hand of a moving person with the help of displacement-time (S-t) and velocity-time (V-t) graph using a ticker timer. Apparatus required : Ticker timer, G-clamp, paper tape, plug key, metre scale, A.C. mains supply, step down transformer 6V, graph paper etc. studied. It is therefore motion of the hand and not the motion of the student because hand pulls the tape. But if there is no relative motion between the student and his/her hand while moving, that is the student is holding the tape without moving his/her hand then it will be the motion of the student also. Therefore in reality we are studying the motion of the student. Procedure: i) Fix a ticker with the help of a G-Clamp on one end of the table. Switch of A.C. mains is connected to primary of transformer or input and secondary of transformer is connected to the electromagnet through a plug key. (fig.). Student walks holding the paper tape ii) Arrangement of apparatus : Ticker timer is fixed with G-clamp on one end of table. A roll of paper tape is also placed near the ticker timer. While doing experiment, a student holds one end of paper tape in his hand and covers a distance of 2-3 meters as shown in Fig. Allow the current to pass through the electromagnet and check whether it is working properly or not. If ticker is vibrating properly and marks are obtained on the tape while pulling it out then the apparatus is ready for use. iii) Now current is passed through the electromagnet and one student holds the free end of tape with his/her hand and moves 2-3 meters slowly and then stops. We see that dots are marked on the tape as shown in (fig.). From fig. it is clear that initially dots are closer, then they move apart and again come closer. Principle : If the student is having a uniform motion then marks produced on the tape due to the ticker are at equal distances, but if the motion is accelerated then the distance between adjacent marks goes on increasing. A tick can be taken as a unit of time. Suppose distance between 10 marks is s, then s is distance traveled in 10 ticks (Remember 1 tick = 1 50 second). Therefore any mark on the paper tape can be taken as origin and the distance covered by the student in 10,20, 30 ticks ca be calculated and therefore s-t or displacement time graph can be obtained. Now distances traveled in successive intervals of 10 ticks can be found from the tape., By dividing these distances by 10 we can find average velocity. Thus the velocity-time graph can also be obtained. By finding rate of increase of velocity, acceleration can be calculated. In this way while a student is moving, complete motion of his/her hand can be Histogram of tape lengths iv) To study motion of hand, take any point in the beginning as origin (say 0) and after that mark P, Q.R.S. etc. at every 10 th point. Now 102 cut OP, PQ, QR and RS…. Parts carefully from the tape. Fix these parts on the left hand page of your note book as shown in fig. Base of all the parts should be in the same straight line of the histogram. Graph : (i) The histogram shows distribution of distance in covered 10 ticks. This distance first increases and then decreases. v) Now distances OP, OQ, OR… are measured with a meter scale and then written in a tabular form .The displacement time graph can be plotted with these observations. vi) Now average velocity is calculated in the intervals: OP, PQ, RS etc. Average velocity can be taken as instantaneous velocity at exactly the middle of the time interval. For example, if PQ=7cm and t=10 tick, then average velocity of hand =7/10= 0.7cm/tick. Here 0.7cm/tick can be taken as instantaneous velocity at t=15 tick which is mid point of 10+20 tick interval. Similarly, velocity can be found at different times and recorded in the observation table. Velocity-time graph can be plotted from these observations. From observation table. A distance S, is plotted on Y-axis and time t on X-axis after taking a suitable scale on a graph paper. The graph so drawn is known as S-t graph of the motion of hand and is shown in fig. S t graph (ii) From observation table B, instantaneous velocity V is plotted on Y-axis and time t on X-axis. This graph is known as V-t graph of motion of hand and is shown in fig. V Observations (A) For displacement-time or (s-t) graph S.No. Time (tick) distance (S) (cm) 1. 10 OP = ........... 2. 20 OQ = ........... 3. 30 OR = ........... 4. 40 OS = ........... 5. 50 OT = ........... 6. 60 OU = ........... (B) For velocity time graph S. Time interval Displacement Average ∆s Velocity No. ∆t ∆s (tick) (cm) ∆t Time Instantaneous (tick) Velocity ∆ S / ∆t V=∆ (cm/tick) 1. 2. 3. 4. 5. 6. 5 10 25 35 45 55 0 10=10 10 20=10 20 30=10 30 40=10 40 50=10 50 60=10 OP = ........... PQ = ........... QR = ........... RS = ........... ST = ........... TU = ........... t Variation of instantaneous velocity with time Result : S-t graph of a moving student’s motion of hand is plotted in fig. and the V-t graph for the same is shown in Fig. From fig. it is clear that displacement increases continuously with time. Initially rate of increase is large but later on it decreases and in the end it becomes zero. From fig. it is clear that initially velocity of hand increases, then in the midele it becomes almost constant and towards the end it decreases rapidly. Dotted portion shows that initial and final velocity should be zero. Precautions : i) During the whole experiment, the tape should remain taut, therefore hand should not move, relative to the body, or relative velocity between hand and the student should be zero. 103 ii) As time interval between two consecutive dots is only 1/50 second, therefore distance between at least 10 dots should be measured, so that magnitude of distance is large and error in measuring distance is less. iii) The surface which touches paper tape should be smooth or frictionless so that no part of kinetic energy of the motion of hand is wasted in overcoming force of friction. iv) Periodic time of ticker should be noted properly and tick should be changed to second and velocity should be found in cm/second or m/second. v) Initially motion of hand is a little irregular, so origin should be selected after leaving a few dots in the beginning. vi) Motion of paper tape should be continuous till the end for one experiment. Paper tape should not be stopped and started in the course of the motion of the student. vii) As the tape is pulled, it must remain horizontal. Sources of error : i) It has been assumed that average velocity at the middle of time-interval is instantaneous velocity at that movement, which may not be exactly true. Smaller the interval, more correct will be this assumption. But too small tick interval will increase errors in measurement. Therefore tick interval should neither be too small not too large. Hence instantaneous velocity can be considered as only approximately correct. ii) Motion of tape does not exactly indicate motion of hand as some friction is present between the paper tape and spool, and also between needle and carbon tape. iii) During experiment tape becomes slightly loose. Therefore time interval between two equally spaced points does not remain same always. Discussion Motion of hands of different persons will be different. Therefore different kinds of graphs are obtained. This is why the shape of V-t graph is not content. VIVA VOCE QUESTIONS 1. What do you understand by displacement of a moving object? 2. What is meant by instantaneous velocity? 3. 4. 5. 6. 7. 8. 9. 10. 1. 2. 3. 4. 5. 6. 7. What is the relation between instantaneous velocity and average velocity? State two differences between displacement and distance. What is ticker timer? Where is the ticker in this experiment? What is a tick? What is the magnitude of tick. What kind of graph is obtained in this experiment in case of a uniformly accelerated motion? How can you find instantaneous velocity from S-t graph? If all points on the paper tape are equally spaced, what kind of motion is possessed by the hand? Why? ANSWER Displacement is change in position of the moving object in definite direction. Velocity of a particle at any instant of time is known as its instantaneous velocity. They are taken as same in the present case, but they may or may not be same. (i) Distance is a scalar but displacement is a vector quantity. (ii) Distance is the length of path cove red by moving object in a given time-interval but displacement is change in position in a given direction in the given time interval. Ticker-timer is a device to measure distance in very small time intervals. Ticker is a steel strip above the carbon disc. It has a needle attached to its one end and its other end is between the pole pieces of an electromagnet. It is called a ticker because it vibrates. The frequency of its vibrations is 50 Hz. Periodic time of ticker or time-interval between two successive dots of paper tape is called a tick. Magnitude of 1 tick = 1 50 second because ticker completes 50 vibrations in 1 second. 8. 9. From s = ut + 1 2 at graph between s & t is 2 parabolic. At any time t a tangent is drawn to the S-t graph. Slope of the tangent gives instantaneous velocity of the object at that time. 104 PROJECT 4 B CONSERVATION OF ENERGY AND MOTION UNDER GRAVITY 1 Introduction : In this chapter we will describe two experiments which can be done with the help of ticker timer. In one experiment law of energy conservation will be verified for a freely falling body and in the other experiment value of acceleration due to gravity will be calculated for the same body. Therefore, to understand these experiments clearly, we should have background knowledge of i) construction of ticker timer its working and principle ii) energy and energy conservation principle iii) laws of motion for a freely falling body under gravity. Ticker timer has already been described and acceleration due to gravity has been described. Here we will describe only the conservation of energy and motion under gravity in brief. 2 Energy : The ability of a body to do work is called energy. Measurement of energy : Energy of an object is measured by the amount of work done to bring the object to a new state. Forms of Energy : It has many forms, for example, mechanical energy, heat energy, light energy, magnetic energy, electrical energy, sound energy, chemical energy, nuclear energy, solar energy etc. Mechanical Energy : Energy produced in due to mechanical work is known as mechanical energy. It is of two kinds (i) kinetic energy (ii) potential energy. Kinetic Energy : It arises due to the motion of a body, when a bullet hits a target it moves inside the target and produces heat. Energy required for production of heat is derived from its K.E. Because part of K.E. is used in overcoming friction. Formula for K.E. : If an object of mass m is moving with a velocity v, it has kinetic energy K= 1 mv 2 2 Momentum and Kinetic energy : If an object has mass m and momentum and kinetic energy as p and K, respectively then K= P2 1 mv 2 = or P = 2 mK 2 2m Work energy therom : Work done by a given force on a given object, is equal to the change in the kinetic energy of the object. Potential energy : It is the energy due to position or configuration of body. Forms of potential energy – Gravitational potential energy, electrostatic potential energy, magnetic potential energy etc. Gravitational potential energy arises due to increase in height of an object from surface of earth. Measurement of potential energy : P.E. is measured by the amount of work done in bringing the object from normal configuration to that configuration or position. If an object of mass m is situated at a height h from surface of earth then its potential energy is mgh where g is acceleration due to gravity. In this formula h << R where R is radius of earth. Also at surface of earth, potential energy is assumed to be zero. Conversion of gravitational potential energy to Kinetic energy: When an object is dropped from a height, its velocity and hence its kinetic energy increases. With the fall its height from earth’s surface decreases, hence its potential energy also decreases. In reality, the gravitational potential energy of the object changes to its kinetic energy. 105 Let an object of mass m be dropped from a height h. At point A, K.E. = 0. Gravitational potential energy = mgh Total energy = 0 + mgh = mgh body A V=0 B V=V1 h kinetic energy, total energy remains constant. Law of conservation of energy – Total energy of an isolated system always remains constant. Energy can neither be created noe destroyed, only it can be transformed from one form to another. Therefore, total energy of universe always remains constant. We’ve seen in above example that total energy of a freely falling object always remains constant. As the object moves down, its potential energy changes to kinetic energy. At surface of earth its kinetic energy= potential energy at the point from where it was dropped. Total energy gy (h–x) ii) At point B- When object has fallen through a distance x then height of object from earth’s surface= h-x and potential energy at B=mg (h-x). 1 2 1 2 energy= mv12 But v = ∴ √2gx K.E.= mx2gx = mgx then kinetic energy = 1 mv 12 = mgx 2 From A to B decrease in P.E. = increase in its kinetic energy and Total energy = mg (h-x) + mgx = mgh Hence we can say that energy has been transformed from potential energy to kinetic energy. Total energy remains constant. (iii) At point C- On surface of earth, potential energy = 0. If velocity of object at C = V 2 then V 2 = 2gh Therefore K.E. of object = 1 m v22 2 1 = m x 2 gh = mgh 2 Therefore when object falls through a height h its entire gravitational potential energy changes to ia nt te Po Conversion of gravitational potential energy to kinetic energy. l Energy ground If the velocity of the object at point B+V1 then kinetic en er V=V2 Ki e n neti er c gy Total energy Conservation of Energy Law of energy conservation for a freely falling ∴ object can be shown graphically as in (fig.) Unit of energy – Energy is a scalar quantity. Its S.I. unit is joule In C.G.S. system its unit is erg. 1 joule = 107 erg. Dimensions of energy : It has same dimension as work i.e. [ML2 T–2] 3 Motion under gravity : Motion of a freely falling body is said to be motion under gravity. It is a case of uniform acceleration in a vertical plane. It has acceleration equal to acceleration due to gravity (g). If u= o then i) V = u + gt= gt ii) S = ut + iii) V2 = u2 + 2gh = 2gh 1 1 gt 2 = gt2 = h 2 2 Here the object covers a height h in time t, and velocity of object at time t = v From eq. (i) it is clear that if a graph is plotted between v and t, then a straight line inclined to time axis will be obtained. Slope of straight line= g as shown in the fig. below : 106 Procedure : V tan θ = g › O (i) Find the mass of given object with the help of a balance. It should be of the order of 100 gram. (ii) Now ticker timer is fixed on one end of stool by a clamp and paper tape is allowed to move over glass rod in such a way that its other end hangs down without any difficulty. (iii) A.C. is allowed to pass through electromagnet of ticker timer and the object is slowly allowed to fall. On tape dots start appearing. (iv) As the object moves down, its position after a definite time interval of 1/50 second is marked on the tape. The distance between successive dots goes on increasing. (v) As the body reaches the ground, ticker timer is stopped. (vi) Let the last do t be E. Cut the tape at a point slightly above E and spread the entire tape on a table. (vii) Leave initial 3-4 dots on the tape as they are very close to each other and it is difficult to calculate velocity from them. Now count the number of points on the rest of the tape. Let the no. of remaining points on tape be 30. Now select 6 points on the tape let them be P, Q, R, S, T, U. Time interval between them will be nearly same. Therefore number of points between them will be nearly same. It should be noted that distance between two successive points will go on increasing. t → v-t graph of a freely falling body. PROJECT 4.1 (B) Aim: To study conservation of energy of a freely falling body with the help of a ticker timer. Apparatus required : Ticker timer, paper tape, Gclamp, pulley or glass rod over which the tape may move, a rubber ball of about 100 gram or any other object, cello tape, metre scale etc. Experimental arrangement Description of apparatus : Apparatus to be used is shown in fig. A stool is placed on one end of a strong table and by means of clamp ticker timer is fixed to stool. On one end of paper tape, rubber ball is fixed tightly with cello tape and other end of paper tape is made to pass over glass rod or pulleys kept on stool. Glass rod should also be strongly clamped to stool, so that when object falls and tape moves, the tape may not slacken. Principle : When an object falls freely under gravity its P.E. decreases, whereas K.E. increases, but throughout its path sum of P.E. and K.E. remains constant. In other words total energy of a freely falling body is always conserved. Therefore we can say that a freely falling object follows the law of conservation of energy. If a freely falling object has mass m and velocity v at height h then its total energy = K.E. + P.E. = 1 mv2 + mgh. 2 (viii) Determine the distance between P and the last dot with the help of a metre scale. It is equal to the instantaneous height of the object above earth’s surface. Similarly find distances of Q, R, S, T and U from last point E. These distances represent distance of the object corresponding to these points from earth’s surface. Therefore potential energy at these heights can be calculated. Now we want to calculate velocity at height h corresponding to points P, Q, R, S, T and U. We will make use of formula. V= Dis tan ce (S) Time (t ) 107 Therefore s and t should be known to us. They can be determined by the following method. Take n number of points on left side of point P and n no. of points on right side of point P. (n can have value 2, 3 or 4). Measure distance between these (2n+1) points with the help of a metre scale. Let it be s. Time taken to travel this distance t= 2n x 1/50 second. Therefore average velocity at height h corresponding to point. P= s s 1 = • t 2 n 50 Similarly calculate average velocity at points Q, R……………. & U. From these values kinetic energy ‰ mv2 of the object can be calculated. Therefore total energy at eny height h can be calculated. ET = (Potential energy + Kinetic energy) we shall see that at any height h value of total energy remains constant. It verifies the law of conservation of energy/ (x) a freely falling body. Precautions : 1. Ticker timer should be fixed at one edge of the stool. 2. Paper-tape should pass over smooth surface of glass-rod so that tape does no get turned down and it should hang vertically downward. 3. As the object is falling freely, no drag should act on the tape. 4. First the tape should be pressed with thumb, as soon as the ticker timer is started, it should be released. 5. As soon as the object touches the ground note down that point E on the tape. Any other point after E should not be considered. 6. To find velocity last point U should be at least 3 points before E. 7. While calculating energy, all quantities should be taken either in C.G.S. or S.I. Units. Here we have calculated energy in ergs. It can be converted to joules according to the relation 1 erg = 10–7 joule. 8. Paper tape should not touch stool or table anywhere. Record all observation and calculations in a tabular form. Observations and Calculations Periodic time of ticker timer T = 1/50 = 0.02 second Mass of object m = ……… gram Sources of error : Frictional forces cannot be removed completely between tape and the timer. Therefore part of energy of falling object is used in overcoming force of friction. Force of friction changes Value of n for calculations = ………… Acceleration due to gravity g = 981 cms–2 S. No. Mark Height of object h (cm) Distance between (2n+1) Points S(cm) Time interval t=2nT (second) Result : Within experimental errors, at every point total energy of a freely falling object remains constant. It proves that law of conservation of energy is applicable for At height h velocity of object v=st cm/ second Kinetic energy of object Ek=‰mv2 (rg) Potential energy of object Ep=mgh (erg) Total energy of object Ek+Ep (erg) with velocity of tape therefore it cannot be easily corrected for. Discussion : In this experiment, average velocity is calculated at every h, whereas instantaneous velocity 108 should be calculated. With reference to this see answer (18). depend on direction. 8. Heat energy, light energy and electrical at energy. Viva Voce Questions 1. What are you doing ? 9. Mechanical energy. 2. How do the potential and kinetic energies of a freely falling object change. Where is potential energy maximum and where is kinetic energy maximum? 13. Potential energy = mgh= 2x3x9.8=58.8 J 14. 2 Kinetic energy = ‰m U2=‰x2x5 =25J 15. No. 3. What is energy? What is its S.I. unit? 16. 4. Write dimensions of energy. 5. What is energy? Define it. 6. What kind of quantity is energy? Scalar or vector? Average velocity is calculated. As velocity of the object is continuously increasing, it is not possible to find velocity at a given instant, but average velocity can be calculated in a given interval. 17. Yes, error can creep in. Less the time interval closer will be average velocity to instantaneous velocity. But on decreasing time interval, distance will also be decreased. Therefore error in measurement of distance (s) may increase. 7. What is law of conservation of energy? 8. State names of any three forms of energy? 9. Which form of energy is conserved in this experiment? 10. What is mechanical energy? What are its different kinds? 18. To reduce this error we should i) keep the value of n less (about 3) 11. What is potential energy? ii) 12. What is kinetic energy? 13. An object of mass 2 kg is placed on top of a house Height of roof top is 3 metere. What is potential energy of the object? take origin of time at initial point (t=)), count number of points till P, Q, R, S, T and U. These quantities tell the time taken to travel these distances. In this way, magnitudes o velocity v at different times can be calculated. 14. A ball of mass 2 kg is moving with velcity of 5ms–1. What is its kinetic energy? 15. Can we calculate instantaneous velocity of the object at height h in this experiment? 16. Which velocity(instantaneous or average) do you calculate? Why? 17. Does it cause any error? 18. What should be done to reduce this error? Answer 2. Potential energy is maximum at the highest point and kinetic energy is maximum on earth’s surface. 5. If force of 1 dyne acting on a body displaces it by 1 cm then work done by the force is 1 erg. 6. Energy is a scalar quantity because it does not With this data, v-t graph can be obtained. This graph should be approximated to nearest straight line. From this graph calculate velocity of object at P, Q, R, S, T and U points. If n is less then velocity obtained from straight line will be closer to instantaneous velocity. Project 4.2 (B) Aim : To find acceleration of a freely falling object with ticker timer. Apparatus required : A ball or any object of mass about 100 gram, metre scale, ticker timer, tape, glass rod over which tape can be passed, paper tape cello tape, a high stool which can be placed on table. GClamp, graph paper etc. Principle : When an object is falling freely, an acceleration acts on it due to gravitational force of attraction of earth. The acceleration of a freely falling object does not depend on mass, shape and size etc. 109 of the object. All freely falling objects fall with same acceleration. This acceleration is know as acceleration due to gravity and it is denoted by g. Although value of g is slightly different at different places but it can be taken as 9.8cm–2. This value is standard value of g at a latitude of 45º at sea-level. than 30, then we can mark 5th, 10th, 15th, 20th, 25th and 30th point from the initial point. 9. Suppose we have to find velocity at the point P. Select 3 points before P and 3 points after? Procedure : 1. Check whether ticker timer is working properly or not. If it is not vibrating properly, get it repaired. 2. Keep stool on table as shown in fig. and with the help of G-clamp fix it at one edge of stool. Paper tape after passing over glass rod, should stand vertically erect. Glass rod reduces friction and prevents wear and tear of tape. 4. 5. 6. 7. 8. One end of tape is passed through ticker-timer and required length of tape (slightly more than height of stool above earth’s surface) is pulled from spool and cut. It is placed straight near the spool so that when object falls down then tape should move into timer without friction and from there should pass over glass-rod then fall down. On other end of the tape fix the object with a cello tape. Object is suspended at the level of stool. Tape is pressed with thumb on top of stool so that object remains stationary at its place. Now electricity is passed through the electromagnet of timer and when the object starts falling down, thumb is removed from tape. ∴ velocity at P = = Dis tan ces between 6 po int s Time int erval between 6 po int s S AB = 6 • (1 / 50) 0.12 If on two sides of fixed point P we select n points and distance between them is S then velocity at point, P, V=S/(2nxT), Where T= Periodic time of timer= (1/50) Second. 10. Similarly find velocities at points Q, R, S, T, and U. 11. To find time taken to reach points Q, R, S, T and U assume a point to be origin at time t=0 count total no. of points up to these points. Let them be n1, n2………… then time upto P = n1 x 1/50, time upto Q= n2 x 1/50 second… etc. Here time interval between any two consecutive dots or points is 1/50 second which is equal to periodic time of the ticker timer. 12. Record your observations in a tabular form. 13. Plot v-t graph by taking velocity along Y-axis and time along X-axis. Graph is a straight line, whose slope tells us the acceleration of the object. Observations and Calculation As object reaches the ground level stop the ticker timer. 1. Periodic time of timer T= (1/50) Second. 2. Value of n for calculation of velocity = 3. Now spread tape on the table. It is seen that initially points are close by but later on distance between any two consecutive points increases. Graph : Leave a few initial points as they are very close to each other since with them it is difficult to do calculations. Mark 6 point on tape in such a way that number of dot between two of them is nearly equal. Let these points be P, Q, R, S, T and U. These points should be at nearly same time interval. For example, if total number of points on tape is little more V ems –1 → 3. Calculation of velocity – Assume the number of points on the tape. O K H D C E G t (second) → Fig. v – t graph HK= ………. Cm/second 110 S. No. Serial no. of selected point (Serial No. of first point is zero) m Time taken by object to reach this point t= mT second EG= ………. Second Distance between (2n+1) points at point m S (cm) Velocity at point m V= S/t (cm/sec.) slope of v-t graph gives accurate value of g. Viva Voce Questions acceleration due to gravity or g = slope of the graph. = HK/EG = ……….. cm/second2 1. What is the aim of your experiment ? = ……………………. Ms–2 2. Why did you say, “In this laboratory”? From straight line graph it is clear that acceleration of the object is the same throughout. 3. Is there a special name for this acceleration? 4. What is its value? 5. Does value of this acceleration depend on mass and size of the falling object. Body = ……………… metre/second2 6. What is its value at the centre of earth? standard value of g = ………….. (from tables) 7. How is g related to weight of an object? 8. What is weight? What is weight of an object at the centre of earth? = .......................... % 9. What is the relation between g and G? Precautions : 10. State SI unit of g and its dimensions. 1. 11. What is weight of an object placed in a freely falling lift? 12. If earth were not rotating about its axis, what effect would it have on weights of an object placed on earth’s surface? Result : PlaceValue of g of a freely falling % error = Experimental value − s tan dard value • 100 S tan dard value The first 2-3 points in the beginning of the tape should be left because they are very close to each other. 2. The tape, after passing over the glass rod, should hang vertically down. 3. When object falls down, the tape should not become loose. 13. What kind of motion does a freely falling object possess? 4. The tape should not touch stool or table anywhere. 14. What is weight of an object in an artificial satellite of earth? 15. Why is weight of an object zero in an artificial satellite of earth? 16. Why is weight of an object not zero on moon when that is also a satellite of earth. 17. In this experiment a graph is plotted between S and t, what kind of graph is obtained? Sources of error : Same as in experiment No. 5. Discussion : The value of g is obtained from the slope of V-t graph. As the object starts the motion from rest, v=gt. The velocity of object is v after time t from the instant it started falling, or v is instantaneous velocity. If n=3 or so then value of v will be quite accurate. Some errors are also removed by plotting a graph and drawing a straight line. Hence 111 ANSWERS 1. 2. To find acceleration of a freely falling object in this laboratory using a ticker timer. Because value of acceleration changes when we move from one place to another on earth’s surface. 3. Yes, this acceleration is known as “acceleration due to gravity”. 4. Its value is 9.81 m/sec2, at 45º latitude at sealevel. 5. No, value of this acceleration is same for all freely falling objects. 6. Zero. 7. Wt. of object= Mass x acceleration due to gravity. (W = mg) 8. Wt of an object is gravitational force of attraction between object and earth. At centre of earth, wt of object =. 9. G is universal constant of gravitation. g= GM/R2 where M is mass of earth and R is radius of earth. 10. S.I unit of g is ms-2 and dimensions are (MºLT –2). 11. Zero. 12. If earth were not rotating, weight of the object would have been more. 13. Motion of a freely falling object in a vertical plane is uniformly accelerated motion. 14. Zero. 15. In an artificial satellite, apparent weight of an object is zero, because satellite is moving in a circular orbit. Therefore, gravitational force of attraction between body and earth is used in providing necessary centripetal force of attraction. 16. Moon is a very big natural satellite of earth. Although apparent weight of an object on moon due to earth is zero but object experiences a gravitational force due to heavy mass of moon. Therefore, object has weight on moon due to moon’s attraction. 17. S= 1 2 gt graph is a parabola. 2