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1 Problem File SIMPLE HARMONIC MOTION EXERCISE –I POLARISER 1. A simple harmonic motion is represented by F(t) = 10 sin (20t + 0.5). Write down its amplitude, anugular frequency, frequency, time period and intial phase, if displacement is measured in m and time in s. 2. A particle performs SHM between two points 0.2 m apart with a period 10 s. How far will the particle be in 1s, after leaving the mid-position between two points. 3. A particle is executing simple harmonic motion according to the equation x 5sin t where x is in cm. How long will the particle take to move from the position of equilibrium to the position of maximum displacement. 4. An object performs simple harmonic motion of amplitude 5 cm and period 4s. If timing is started when the object is at the centre of an oscillation (i.e. x = 0), calculate (a) the frequency of the oscillation, (b) the displacement 0.5 s after the start. (c) the maximum acceleration of the system and (d) the velocity at a displacement of 3 cm 5. A particle that is at the origin of co-ordinates at t = 0 vibrates about the origin along the y-axis with a frequency of 20 Hz and an amplitude of 3.0 cm. Given its equation of motion. 6. A body executes 40 oscillations per minute. Its maximum speed is 36 cm s–1. Calculate the amplitude of oscillation. 7. A particle executes a SHM of period second and amplitude 2 cm. Find its acceleration when it is at the maximum displacement from the mean position. 8. The displacement x (in meters) of an oscillating particles varies with time t (in second) according to the equation x 0.02cos 0.5t . Calculate (a) amplitude of oscillation (b) time period of oscillation (c) 3 maximum velocity of particle (d) maximum acceleration of particle. 9. A particle executes SHM of period 4 seconds and amplitude 0.02 m. Determine (i) its maximum speed, (ii) speed at displacement 0.01 m, (iii) value of acceleration at the turning point, and (iv) value of acceleration at displacement 0.01 m. 10. A particle executes SHM of amplitude 25 cm and time period 3s. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position ? 11. The vertical motion of a huge piston in a machine is approximately simple harmonic with a frequency of 0.50/s. A block of 10 kg is placed on the piston. What is the maximum amplitude of the piston's SHM for the block and the piston to remain together ? 12. A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5×10–2 m. What must be the least period of these oscillations so that the objects is not detached from the platform ? Take g = 10 m s–2. 13. In a certain engine, a piston undergoes vertical SHM with amplitude 7 cm. A washer rests on top of the piston. As the motor speed is slowly increased, at what frequency will the washer no longer stay in contact with the piston ? 14. A particle is executing SHM with an amplitude of 4 cm and time period 12 second. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is t1. The time taken from this displaced t1 position of 2 cm to reach the extreme position is t2. Calculate the ratio t . 2 15. In what time after its motion begins, will a particle oscillating according to the equation x = 7 sin 0.5 t move from the mean position to the maximum displacement ? 16. A particle executing simple harmonic motion along a straight line has a speed of 4 m s–1 when at a distanne of 3 m from the mean postion and 3 m s–1 when at a distance of 4 m from the mean position. Calculate the time it takes to travel 2.5 m from the positive extremity of the oscillation. Take the value of as 3 only.. 17. A small child bounces up and down while staying in contact with a trampoline. She undergoes SHM and repeats her motion every 1.75 s. The height of each bounce above the equilibrium position of the trampoline is 35.0 cm. (a) What is the amplitude and angular frequency of the child's motion ? (b) Write an equation of motion y(t) for the child's motion if timing is to begin at the highest point of the motion. (c) What is the child's position at t = 4.50 s ? 18. A particle of mass 0.2 kg execute simple harmonic motion of amplitude 2 cm and period 6 s. Find (i) total mechanical energy at any instant (ii) kinetic and potential energies when the displacement is 1 cm. PHYSICS BY ABHISHEK KUMAR | Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in 2 SIMPLE HARMONIC MOTION 19. A metal cylinder of mass 4kg vibrates harmonically with an amplitude of 0.25 m and period of 2 second. Calculate (i) the maximum acceleration and its acceleration when y = 0.15 m.(ii) its maximum restoring force and restoring force when y = 0.15 m (iii) total energy. 20. A particle executes SHM of amplitude a. (i) At what distance from the mean position is its kinetic energy equal to potential energy (ii) At what points is its speed half the maximum speed. 21. A 2kg body panel of a car oscillates with a frequency of 2 Hz and amplitude 2.5 cm. If the oscillations are assumed to be simple harmonic and undamped, calculate (a) the maximum velocity of the panel (b) the total energy f the panel (c) the maximum potential energy of the panel, and (d) the kinetic energy of the panel 1.0 cm from its equlibrium position. 22. A body of mass 9 kg is oscillating on a spring of force constant 100 N m –1. Deduce angular frequency. 23. A massless spring steel is clamped at its lower end. A 2 kg ball is fastened to its top end. A force of 8 N is required to displace the ball 20 cm to one side as shown. Assume the system to undergo simple harmonic motion when released. Find (a) the force constant of the spring and (b) the period with which the ball will vibrate back and forth. 20 cm 8N 24. A spring compressed by 0.1 m develops a restoring force of 10 N, A body of mass 4 kg is placed on it. Deduce (i) the force constant of the spring (ii) the depression of the spring under the weight of the body (iii) the period of oscillation if the body is disturbed. g = 10 m s–2. 25. A spring of force constant 1200 N m –1 is mounted on a horizontal table as shown. A mass of 3.0 kg is attached to the free end of the spring. The mass is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. What is (a) the speed of the mass when the spring is compressed by 1.0 cm. (b) potential energy of the mass when it momentarily comes to rest. (c) total energy of the oscillating mass ? m 26. A 3 g particles at the end of a spring moves according to the equation y = (0.75 sin 63t) cm where t is in second. Find (a) the amplitude (b) frequency of motion (c) position at t = 0.020 s and (d) the spring constant. 27. A 0.3 kg mass is connected to a system of two identical springs as shown in Fig. Determine the period of oscillation. k=20 Nm –1 k=20 Nm –1 0.3kg 28. Fig shows the position function for a 0.500 kg mass executing SHM at the end of a spring. (a) What is the amplitude, period and angular frequency for this SHM ? (b) What is the spring constant k ? (c) Find the position x(t) for this mass. (d) What is the mass's velocity and acceleration at t = 5.00 s ? 8 6 4 2 0 –2 –4 –6 0 2 4 6 8 29. What is the length of a pendulum for which time period is 1 s at a place where g = 0.8 m s–2 ? 30. The time taken by a simple pendulum to perform 100 vibrations is 8 minute 9 second in Bombay and 8 minute 20 second in Pune. Calculate the ratio of accelerations due to gravity in Bombay and Pune. Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in | PHYSICS BY ABHISHEK KUMAR 3 Problem File 31. A vertical U-tube of uniform cross-section contains water upto a height of 2.45 vm. When the water on one side is depressed and then released, its up and down motion in the tube is simple harmonic motion. Show that the time period is equal to seconds. 10 32. A mass suspended by a spring oscillates in 100 cm3 of water. The force constant of the spring is 9800 N m–1. The spring is stretched by 0.10 m and left to oscillate. Find the rise in temperature of water when the oscillations stop. Assume that the whole of the energy of oscillation is used up in increasing the temperature of water. 33. The simple harmonic motion of a particle is given by the equation y 3 sin t 4 cos t . Find its amplitude. PHYSICS BY ABHISHEK KUMAR | Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in 4 SIMPLE HARMONIC MOTION EXERCISE –II 10. k1 (a) k2 11. m 3. 4. 5. k1 k 2 (a) 2 m m (b) 2 (k k ) 1 2 m(k1k 2 ) (c) 2 (k k ) 1 2 (d) 2 6. 7. 8. m(k1 k 2 ) (k1k 2 ) 12. 6sin 3t and Y2 = 5(sin3 t+ 3 cos3 t). 4 Their amplitudes are in the ratio of :– (a) 2 : 1 (b) 3 : 1 (c) 1 : 3 (d) 1 : 4 A mass M attached to a spring oscillates with a period of 2 seconds. If the mass is increased by 2 kg the period increases by 1 second. The initial mass M will be :– (a) 1.6 kg (b) 1 kg (c) 1.5 kg (d) 2 kg The ratio of kinetic enegy at mean position to the potential enegy when the displacement is half of the amplitude is :– 2 (b) 3 4 (c) 3 1 (d) 2 If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression 4v2 = 25 – x 2, then the time period is :– (a) (b) 2 (c) 4 (d) 6 In a simple pendulum at mean position :– (a) K.E. is maximum and PE is minimum (b) K.E. is minimum and PE is maximum (c) Both P.E. and K.E. are maximum (d) Both PE and KE are minimum Maximum velocity in SHM is vm. The average velocity during the motion from one extreme point to the other extreme point will be :– (a) vm 2 (b) 2 vm (c) 4 vm (d) vm 4 5 3 (b) 3 5 3 2 A (b) 2A 3 A 4 (c) (d) 25 9 (d) 3 A 2 Two blocks of mass m1 and m2 are kept on a smooth horizontal table as shown in Fig. Block of mass m1 but not m2 is fastened to the spring. If now both the blocks are pushed to the left so that the spring is compressed a distance d. The amplitude of oscillation of block of mass m1, after the system is released is :– m1 13. 16 9 (c) What will be the displacement of a particle in SHM when its velocit is half the maximum velocity (A = amplitude) (a) Two SHMs are represented by the equations Y1 = 4 (a) 1 When a particle oscillates simple harmonically,its kinetic energy varies periodically. If frequency of the particle is n, the frequency of the kinetic energy is :– (a) n/2 (b) n (c) 2n (d) 4n A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released, so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. The ratio of m/M is:– m2 m1 (a) d m m 1 2 m2 (b) d m m 1 2 2m2 (c) d m m 1 2 2m1 (d) d m m 1 2 Displacement-time graph of a particle executing SHM is shown. The correct corresponding force-time graph of the particle is :– y Time y (a) x y (b) Force 9. Force 2. A simple harmonic motion (SHM) has an amplitude A and time period T. The time required by it to travel frm x = A to x = A/2 is :– (a) T/6 (b) T/4 (c) T/3 (d) T/2 A mass m is suspended from two springs of spring constant k1 and k2 as shown in the Fig. The time period of vertical oscillations of the mass will :– Force 1. ANALYSER x Time x Time Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in | PHYSICS BY ABHISHEK KUMAR 5 Problem File (d) x x Time 14. Time 3E and its displacement y is given by :– 4 15. 16. (a) A particle starts executing SHM of amplitude a and total energy E. At the instant, its kinetic energy is 20. a a a 3 (b) y (c) y (d)y = a 2 2 2 The periodic time of a mass suspended by a spring (force constant k) is T. If the spring is cut in three equal pieces, the force constant of each part and the periodic time if the same mas is suspended from one piece are :– (a) y (a) k, T / 3 (b) 3k, T (c) 3k, 3T (d) 3k, T / 3 21. 22. 1 v 2 v1 x 2 x1 2 (a) v 22 v12 23. 1 v 2 x1 2 v1x 2 2 2 (b) v 22 v12 17. 1 m 2k (b) 2 mgsin (c) 2 2k 18. 24. On smooth inclined plane, a body of mass m is attached between two massless springs. The other ends of the springs are fixed to firm supports. If each spring has force constant k, the period of oscillation of the body is :– (a) 2 (b) 2mgsin (d) 2 k 3 T 2 4 (c) 2 2 (d) 2 4 A particle is executing SHM with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is :– (b) 5 cm (c) 2 3 cm (d) 2 5 cm The period of a simple pendulum is doubled, when :– (a) Its length is doubled (b) The mass of the bob is doubled (c) Its length is made four times (d) The mass of the bob and the length of the pendulum are doubled The period of oscillation of a simple pendulumof constant length of earth surface is T. Its period inside a mine is :– (a) Greater than T (b) Less than T (c) Equal to T (d) Cannot be compared A simple pendulum has a time period T in vaccum. Its time period when it is completely immersed in a liquid of density one-eight of the density of material of the bob is :– 7 8 3 8 T T (c) T (d) T (b) 8 5 8 7 A man measures the period of a simple pendulum inside a stationary lift and finds it to be T s. IF the lift accelerates upward with an acceleration g/4, then the period of the pendulum will be :– Pendulum Lift 2m k A person measure the time period of a simple pendulum inside a stationary lift and finds it to be T. If the lift starts accelerating upwards with an acceleration of g/3, the time period of the pendulum will be :– (a) 3 T (b) (a) v1x 2 v 2 x1 2 (d) 2 2 v1 v 2 (c) v v 2 2 1 2 (a) 3 cm x A sin t , represents the equation of a SHM. If displacement of the particle are x 1 and x 2 and velocities are v1 and v2, respectively, then amplitude of SHM is :– v1x 2 The displacement x (in centimeters) of an oscillating particle varies with time t (in seconds) as x 2cos 0.5t . The magnitude of the maximum 3 acceleration of the particle in cms–2 is :– Force Force (c) 19. y y 2T T (c) (d) 2T 5 5 4 What is the velocity of the bob of a simple pendulum at its mean position. If it is able to rise to vertical height of 10 cm (g = 9.8 m/s2) :– (a) T 25. (b) A (c) T / 3 (d) T/3 (a) 2.2 m/s (c) 1.4 ms B (b) 1.8 m/s M (d) 0.6 m/s PHYSICS BY ABHISHEK KUMAR | Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in 6 26. SIMPLE HARMONIC MOTION The time period of a simple pendulum of length L as measured in an elevator descending with acceleration (a) 2 3L (b) g 3L g 3L (c) 2 2g 27. (d) 2 K2 2L 3g Two identical springs of constant K are connected in series and parallel as shown in figure. A mass m is suspended from them. The ratio of their frequencies of vertical oscillations will be :– 31. m (B) K (a) 1 K 2 2 1 1 (b) 2K1 K 2 1 1 (c) 2K K 1 2 2 1 (d) K1 K 2 (a) (A) (a) 2 : 1 (b) 1 : 1 (c) 1 : 2 (d) 4 : 1 As shown in figure, a simple harmonic motino oscillator having identical four springs has time period :– 32. k k k 33. k (a) T 2 (c) T 2 29. m 4k (b) T 2 m k (d) T 2 A mass m = 100 g is attached at the end of a light spring which oscillates on a frictionless horizontal table with an amplitude equal to 0.16 metre and time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement x = –0.16 metre. The expression for the displacement of the mass at any time t is :– (a) x = 0.16 cos( t ) 30. m 2k 2m k s 20 (b) s 10 (c) s 5 (d) 200 s 2 10 ) :– (a) 6.3 m (b) 0.63 m (c) 0.25 m (d) 63 cm Two simple harmonic motions are represented by (a) 7 34. 1 A body of mass 4.9 kg hangs from a spring and oscillates with a period 0.5s. On the removal of the body, the spring shortened by (Take g = 10 ms–2, y1 4 sin(4t ) and 2 resultant amplitude is :– m 1 A weightless spring of length 60 cm and force constant 200 N/m is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of 0.25 kg is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is :– m 28. K1 K1 g is :– 3 (b) 1 y 2 3 cos(4t) . (c) 5 The (d) 2 3 Two simple harmonic motions are represented by y1 5(sin2t 3 cos 2t) y 2 5 sin 2t 4 The ratio of the amplitude of two SHM's is :– (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 1: 3 (b) x = –0.16cos( t ) (c) x = 0.16sin( t + ) (d) x = –0.16sin( t+ ) What will be the force constant of the spring system shown in the figure :– Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in | PHYSICS BY ABHISHEK KUMAR 7 Problem File ANSWER KEY Exercise–I 1. 10m, 20 rad s–1 , 3.18 Hz, 0.314 s, 0.5rad s–2, 2. 58.8 mm (d) ± 6.4 cms–1 5. 3. 4. (a) 0.25 Hz, (b) 3.54 cm (c) –12.8 cm y = (3.0 sin 125.6t) cm 6. 7. 8 cm s–2 9. (i) 0.01 ms–1 , (ii) 2.72 ×10–2 m s–1 (iii) –49.4 × 10–3 ms–2 (iv) –0.0247 m s–2 10. 0.5 s 11. 0.99 m 12. 0.3142 s 14. 1 2 15. 1 s 16. 1 s 17. (a) 35.0 cm, 3.59 cm s–1, (b) y(t) = 35.0 cos[(3.59 rad s–1)t] (c) –31.5 cm 18. (i) 4.38 × 10–5 J (ii) 3.29×10–5 J, 1.10×10–5 J 19. 0.5 s 3.82 cm 8. (a) 0.02 m, (b) 4 s (c) 3.142×10–2 ms–1 (d) 4.94 × 10–2 m s–2 13. above v = 1.88 Hz (i) 0.252ms2 , 0.152ms2 (ii) 2 N, 0.6 2N , (iii) 0.125 2J 20. (i) 21. (a) 0.314 m s–1 , (ii) 0.1 J (c) 0.1 J, (d) 0.083 J 23. (a) 40 N m–1 (b) 1.40 s 24. (i) 100 N m–1, (ii) 0.4 m, (iii) 26. (a) 0.75 m (b) 10.03 Hz, (c) 0.714 cm (d) 11.9 N m–1 27. 0.54 s 28. 1 (a) 6.00 cm, 1.57 rad s–1 (b) 1.23 N m–1 (c) x(t) = (6.00cm) cos rads t 1.23rad (d) –3.24 cm s–1, – 14.0 cm s–2 2 29. 24.8 cm 33. 5 units 2 s 5 22. 3.33 rad s–1 a 2 (ii) ± 3 a 2 25. (a) 0.35 ms–1 (b) 0.24 j (c) 0.24 J 30. 1.0455 31. 32. 0.177ºC Exercise–II 1. A 2. D 3. D 4. A 5. A 6. C 7. A 8. B 9. C 10. C 11. D 12. A 13. B 14. B 15. D 16. B 17. A 18. B 19. C 20. C 21. C 22. A 23. D 24. C 25. C 26. C 27. C 28. C 29. B 30. B 31. A 32. C 33. 34. C A PHYSICS BY ABHISHEK KUMAR | Beside Canara Bank, Bazar Samiti, Patna-16, Mob.: - 8877667519, www.abhishekphysics.in