Download PROJECT

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