1. Dimensions of force constant are (MHT-CET-2003) (a) M 1 L 0 T 2 (b) M 1 L 0 T 2. (c) M 0 L 1 T 2 (d) MLT 2

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1 1. Dimensions of force constant are (MHT-CET-2003) M 1 L 0 T 2 M 1 L 0 T 2 M 0 L 1 T 2 MLT 2 2. A bar magnet is in oscillatory motion its frequency is n magnetic field induction is T when the frequency is doubled due to increase the magnetic field induction. The what is increasing value to induction? B B A particle moves such that its acceleration a is given by a bx, where x is the displacement from equilibrium position and b is a constant. The period of oscillation is (CPMT 91, NCERT 84) 4. A force of 6.4N stretches a vertical spring by 0.1 m. the mass that must be suspended from the spring so that it oscillates with a period of ( /4) sec is (Roorkee 88) 1 kg 10 kg 5. Total energy of a particle executing S.H.M. is proportional to (MHT CET 2002, CPMT 74) Square of the amplitude of the motion Frequency of oscillation Velocity in equilibrium position

2 Displacement form equilibrium position 6. The length of a pendulum is halved. Its energy will (MH- CET-2002) Decreased to half Increased to 2 times Decreased to one fourth Increased to 4 times 7. In simple harmonic motion which statement is wrong. (MHT-CET-2008) A body in S.H.M. its velocity maximum at mean position A body in S.H.M. its K.E. less at extreme position A body in S.H.M. its acceleration more at extreme position its directions away from mean position A body in S.H.M its acceleration less at mean position 8. A simple pendulum is made of a body which is a hollow sphere containing mercury suspended by means of a wire. If a little mercury is drained off, the period of pendulum will (NCERT 72) Remain unchanged Increase Decrease Become erratic 9. Time period of pendulum is 6.28 sec and amplitude of oscillation is 3 cm. Maximum acceleration of pendulum is (MHT-CET-2002) 8 cm/s cm/s 2 3 cm/s cm/s A body in simple harmonic motion. Its time period is 24 sec. at mean position to 4 sec after its velocity is m/sec. then find its

3 path length. (MHT-CET-2008) 48 m 58 m 68 m 78 m 11. If the length of second s pendulum is increased by 2%, how may seconds it will lose per day? (CPMT 92) 3427 sec 3727 sec 3927 sec 864 sec 12. A mass of 1 kg attached to the bottom of a spring ahs a certain frequency of vibration. The following mass has to the added to it in order to reduce the frequency by half (Roorkee 88) 1 kg 3 kg 2 kg 4 kg 13. In S.H.M. path length is 4 cm and maximum acceleration is 2 2 cm/s 2. Time period of motion is (MHT-CET-2002) 2 s 4 s s 1/2 s 14. A magnet, when suspended in an external magnetic field, has period of oscillation of 4s. when it is cut length wise, and suspended in the same magnetic field, the period of vibration will be (MHT-CET-2007) 2 s 8 s 15. A body executes S.H.M. with an amplitude A. At what displacement, from the mean position, the kinetic energy of the body is one fourth of its total energy (CBSE 90)

4 16. M kg weight is suspended from a weightless spring and it has time period T. if now 4M kg weight is suspended from the same spring, the new time period will be (CPMT 79) T 2T 4T 17. The acceleration of particle executing S.H.M. when it is at mean position is (MHT CET 2002) Infinite Varies Maximum Zero 18. A.S.H.M. is represented by x (sin 2 t cos 2 t). Then amplitude of the S.H.M. is (MHT CET 2004) 10 cm 20 cm 5 cm 50 cm 19. W denotes to the total energy of a particle in linear S.H.M. At a point, equidistant from the mean position and extremity of the path of the particle (MH-CET 2001) K.E. of the particle will be w/2 and P.E. will also be w/2 K.E. of the particle will be w/4 and P.E will be w/4 K.E. of the particle will be 3w/4 and P.E. will be w/4 K.E. of the particle will be w/8 and P.E. will be 7w/8 20. In SHM, graph of which of the following is a straight line? (MHT-CET2007) T.E. against displacement

5 P.E. against displacement Acceleration against time Velocity against displacement 21. The potential energy of a particle in S.H.M. at a distance x from the equilibrium position is (MH-CET 99) Zero 22. The velocity of a particle performing simple harmonic motion, when it passes through its mean position is (MHT CET 2002) Infinite Zero Minimum Maximum 23. The time period of a bar magnet in uniform magnetic field is T. It is cut into two halves, by cutting it parallel to its length then the time period of each part in same field is (MHT-CET-2001) T 2T None of these 24. The period of oscillation of a mass M, having from a spring of force constant k is T. When additional mass m is attached to the spring, the period of oscillation becomes 5T/4. m/m = (MHT- CET-2007) 9:16 25:16 25:9 19:9

6 25. The pendulum energy of a particle executing S.H.M. at a distance x from a equilibrium position is proportional to (Roorkee 92) 1. Answer: 2. Answer: 3. Answer: 4. Answer: 5. Answer: 6. Answer: 7. Answer: 8. Answer: 9. Answer: x x 2 x Answer: 11. Answer: 12. Answer: 13. Answer: 14. Answer: 15. Answer: 16. Answer: 17. Answer: 18. Answer: 19. Answer: 20. Answer: 21. Answer: 22. Answer: 23. Answer:

7 24. Answer: 25. Answer: 1. The period of simple pendulum is doubled when (CPMT 74) Its length is doubled Its length is halved The length is made four times Mass of the bob is doubled 2. If the length of a simple pendulum is doubled keeping its amplitude constant its energy will be (MHT CET 2001) Unchanged Doubled Four times Halved 3. If an simple pendulum oscillates with an amplitude of 50 mm and time period of 2s, then its maximum velocity is (MHT- CET-2000) 0.10 m/s 0.16 m/s 0.25 m/s 0.5 m/s 4. Starting from the extreme position, the time taken by an ideal simple pendulum to travel a distance of half of the amplitude is (MHT-CET-2007) T/6 T/12 T/13 T/4

8 5. If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0 ms 2 at any time, the angular frequency of the oscillator is equal to (CBSE 92) 10 rad s rad s rad s 1 1 rad s 1 6. A body is executing S.H.M. when the displacements from the mean position are 4 cm and 5 cm, the corresponding velocities of the body are 10 cm/s and 8 cm/s respectively. The time period of oscillation is (MHT CET 2001, CBSE 91) 5 second 3.14 second 2 second 6.28 second 7. The period of thin magnet is 4 sec. if it is divided into two equal halves then the time period of each part will be (MHT CET 2004) 4 sec 1 sec 2 sec 8 sec 8. The force constant of a wire is K and that of another wire of the same material is 2K. when both the wires are stretched, then work done is (MHT-CET-2000) W W 1 W 2 W 1 W 2 2W 2 W 2 2W The acceleration due to gravity changes from 9.8 m/s 2 to 9.5 m/s 2. To keep the period of pendulum constant, its length must changes by (MHT-CET-2006) 3 m 0.3 m

9 0.3 cm 3 cm 10. A simple harmonic oscillator has an amplitude a and time period T. the time required to travel from x is (CBSE 92) 11. Two springs of constants k 1 and k 2 equal maximum velocities, when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be (MHT-CET-2000) 12. A load of mass 100 gm increases the length of wire by 10 cm. If the system is kept in oscillation, its time period is (MHT-CET- 2006) s 3.14 s s 6.28 s 13. A spring has a force constant K and a mass m is suspended from it. The spring is cut into half and the same a mss is suspended from one of the halves. If the frequency of oscillation in the first case is, then the frequency in the second case will be (CPMT 86) A particle executes S.H.M. with amplitude 0.5 cm and frequency

10 100 Hz. The maximum speed of particle will be (CPMT 84) m/s m/s 0.5 m/s 100 m/s 15. A particle executes simple pendulum harmonic motion of amplitude A. at what distance from the mean position is its kinetic energy to its potential energy? (MHT-CET-1999) 0.51 A 0.61 A 0.71 A 0.81 A 16. A particle of mass m is executing SHM about its mean position. The total energy of the particle at given instant is 17. Two bodies M and N of equal masses are suspended from two separate mass less springs of spring constants k 1 and k 2 respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitudes of M to that of N is (IIT 88) 18. A simple harmonic motion having an amplitude A and time period T is represented by the equation y 5 sin (t 4)m then the values of (A in m) and (T in sec) are (MNR 91) A 5, T 2 A 10, T 1 A 5, T 1 A 10, T If the length of simple pendulum is increased by 44% then what is the change in the time period of the pendulum? (MHT-CET

11 2004) 22 % 20 % 33 % 44 % 20. The maximum velocity and maximum acceleration of a body moving in a simple harmonic oscillator are 2m/s and 4m/s 2 the angular velocity is (MHT-CET-1999) 1 rad/s 2 rad/s 4 rad/s 5 rad/s 21. If a bar magnet of magnetic moment M is kept in a uniform magnetic field B, its time period of oscillation is T. The another magnet of same length and breadth is kept in a same magnetic field. If magnetic moment of new magnet is M/4, then its oscillation time period is (MHT-CET-2006) T 2T T/2 T/4 22. For a particle executing simple harmonic motion, the kinetic energy K is given by K K 0 cos 2 t. the maximum value of potential energy is (CPMT 81) K 0 Zero Not obtainable 23. A body of mass 5 gm is moving at the centre with amplitude of 10 cm. its maximum velocity is 100 cm/s. its velocity will be 50 cm/s at a distance (in cm) (CPMT 76) The unit of force constant is (MHT-CET 99)

12 Nm N/m N/kg Nkg 25. The maximum velocity of a body in S.H.M. is 0.25 m/s and maximum acceleration is 0.75 m/s 2, the period of S.H.M. is second second second second 1. Answer: 2. Answer: 3. Answer: 4. Answer: 5. Answer: 6. Answer: 7. Answer: 8. Answer: 9. Answer: 10. Answer: 11. Answer: 12. Answer: 13. Answer: 14. Answer: 15. Answer: 16. Answer: 17. Answer: 18. Answer: 19. Answer: 20. Answer:

13 21. Answer: 22. Answer: 23. Answer: 24. Answer: 25. Answer:

EXERCISE - 01 NEETIIT.COM. d x dt. = 2 x 2 (4)

EXERCISE - 01 NEETIIT.COM. d x dt. = 2 x 2 (4) 1. The acceleration of a particle executing S.H.M. is (1) Always directed towards the equillibrium position () Always towards the one end (3) Continuously changing in direction () Maximum at the mean position.