The displacement (in mm) of the vibrating cone of a large loudspeaker can be represented by the equation x = 10 cos (150t), where t is the time in s. Which line, A to D, in the table gives the amplitude and frequency of the vibrations.

amplitude/mm frequency/Hz

A 5 10/2

B 10 150

C 10 150/2

D 20 150/2

--

amplitude/mm frequency/Hz

A 5 10/2

B 10 150

C 10 150/2

D 20 150/2

Look at the equation for displacement - A is the term before the 'cos'. In this case '10' so the amplitude is 10

150 corresponds to 2 f

so f = 150/2

Q17. A mechanical system is oscillating at resonance with a constant amplitude. Which one of the following statements is not correct?

A The applied force prevents the amplitude from becoming too large.

B The frequency of the applied force is the same as the natural frequency of oscillation of the system.

C The total energy of the system is constant.

D The amplitude of oscillations depends on the amount of damping.

--

The applied force is what is producing resonance not limiting it - A is false.

Constant amplitude means that there is equilibrium - applied force's expected increase in the system's energy will be balanced by the system's loss of energy. C is tue.

For resonance to occur B could be true - or the applied force could be at a multiple of the frequency - but for constant amplitude to be maintained the applied force would need to be at the frequency - no time for a slow down inbetween applications.

D is true - heavily damped systems have lower amplitude.

Q18. A particle of mass 0.20 kg moves with simple harmonic motion of amplitude 2.0 × 10–

2m.

If the total energy of the particle is 4.0 × 10–5J, what is the time period of the motion?

A – π/4 seconds

B – π/2 seconds

C - π seconds

D - 2π seconds

--

Total energy = max kinetic energy = 1/2 mv 2 at the centre of the path where velocity = maximum

vmax = 2πfa - from data sheet

Total energy =1/2 m (2πfa)2 = 4.0 × 10–5

1/2 x 0.2 (2πf x2.0 × 10–2)2 = 4.0 × 10–5

4π2f2x 4.0 × 10–6 = 4.0 × 10–6

f2 = 1/(4π2)

T2 = 4π2

∴T = 2π - choice D

The graph shows the variation in displacement with time for an object moving with simple harmonic motion.

What is the maximum acceleration of the object?

A 0.025 m s–2

B 00.99 m s–2

C 002.5 m s–2

D 009.8 m s–2

--

Maximum acceleration = (2πf)2A - from the data sheet

From the graph:

A = 10 x 10-2 m

T = 2 s

Now f = 1/T = 0.5 Hz

So, amax = (2π x 0.5)2 x 10 x 10-2

amax = π2 x 10 x 10-2 = 0.987m s–2 - choice B

Q20. Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90 s and the period of Q is 1.95 s.

How many oscillations are made by pendulum Q between two consecutive instants when P and Q move in phase with each other?

A 19

B 38

C 39

D 78

--

The difference in time period is 0.05s - therefore to make up a complete extra swing P will move one more period than Q does in the same time.

Let number of swings of Q = n

Then n x 1.95 = (n+1) x 1.90

1.95n = 1.90n + 1.90

0.05n = 1.90

n = 38 - choice B

Q21. A particle of mass m oscillates in a straight line with simple harmonic motion of constant amplitude. The total energy of the particle is E.

What is the total energy of another particle of mass 2m, oscillating with simple harmonic motion of the same amplitude but double the frequency?

A 2E

B 2E

C 4E

D 8E

--

Total energy = maximum KE = maximum PE

KEmax = 1/2 mvmax2

KEmax = 0.5 x m (2πfA)2

KEmaxis ∝to mf2

therefore the new particle will have energy x 2 (for mass doubling) and x 22 (for frequency doubling)

a multiplication factor of 2 x 4 = 8

∴ the answer is choice D

Q22. When a mass suspended on a spring is displaced, the system oscillates with simple harmonic motion.

Which one of the following statements regarding the energy of the system is incorrect?

A The potential energy has a minimum value when the spring is fully compressed or fully extended.

B The kinetic energy has a maximum value at the equilibrium position.

C The sum of the kinetic and potential energies at any time is constant.

D The potential energy has a maximum value when the mass is at rest.

--

When a mass M attached to a spring X, as shown in Figure 1, is displaced downwards and released it oscillates with time period T.

An identical spring is connected in series and the same mass M is attached, as shown in figure 2.

What is the new time period?

A T/2

B

C

D 2T

--

Putting the springs in series halves k (stretch of the system doubles for a given mass)

T1√k 1= T2√k 2

T2 = T1√(k 1/k 2)

T2 = T√2 - choice C

a In the simple harmonic motion, there is a relation between mass acceleration and

1. the period

2. the speed

3. the displacement

4. the frequency

b In the simple harmonic motion, velocity has a maximum when

1. displacement has a maximum

2. acceleration is zero

3. period has a maximum

4. frequency has a maximum

c In the simple harmonic motion, when the displacement from equilibrium position has a

maximum

1. potential energy has a maximum and kinetic energy a minimum

2. potential energy is a quarter of kinetic energy

3. potential energy has a minimum and kinetic energy a maximum

4. kinetic energy is a quarter of potential energy

d A 10 kg mass oscillates with 20 cm amplitude, hanged from a spring of stiffness constant

100 N/m. When the mass passes trough the equilibrium position, its kinetic energy is

1. 20 J

2. 4 J

3. 2 J

4. 40 J

e A 500 g mass oscillates with amplitude that decreases with time, hanged from a spring of

stiffness constant 125 N/m. If the mass loses the half of its energy in 4s, the relative energy

loss per cycle is

1. 68.9%

2. 89.6%

3. 8.96%

4. 6.89%

3. The displacement of particle performing simple harmonic motion is given by, x 8 sin t 6 cos t, where distance is in cm and time is in second. The amplitude of motion is

(a) 10 cm (b) 14 cm

(c) 2 cm (d) 3.5 cm

1. The length of second’s pendulum on the surface of earth is 1 m.

the length of same pendulum on the surface of moon, where acceleration due to gravity is (1/6)th of the g on the surface of earth is (NCERT 71)

(a) 36 m (b) 1 m

(c) (d)

2. A mass M is suspended from a light spring. If the additional

mass m is added, it displaces the spring by a distance x. now the combined mass will oscillate on the spring with time period equals to (CPMT 89)

(a) (b)

(c) (d) 3. The displacement of particle performing simple harmonic

motion is given by, x 8 sin t 6 cos t, where distance is in cm and time is in second. The amplitude of motion is (MHT-CET-2005)

(a) 10 cm (b) 14 cm

(c) 2 cm (d) 3.5 cm

4. A simple pendulum is set up in a trolley which moves to the

right with an acceleration a on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle with the vertical (CPMT 83)

(a)

(b)

(c)

(d)

5. The angular velocity and the amplitude of a simple pendulum is

‘’ and ‘a’ respectively. At a displacement x from the mean position its kinetic energy is T and potential energy is V, then the ratio of T to V is

(CBSE 91)

(a) (b)

(c) (d) 6. A particle executes S.H.M. of amplitude A. at what distance

from mean position its kinetic energy is equal to its potential energy? (MHT-CET 99)

(a) 0.51 A (b) 0.61 A

(c) 0.71 A (d) 0.81 A

7. A simple pendulum of length l and mass (bob) m is suspended vertically. The string makes an angle with the vertical. The restoring force acting on the pendulum, is (MHT-CET-2005)

(a) mg tan (b) mg sin

(c) mg sin (d) mg cos

8. The mass and diameter of a planet are twice those of earth. the

period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth)

(IIT 73)

(a) (b)

(c) 2 second (d)

9. A second’s pendulum is placed in space laboratory orbiting

around the earth at a height 3R from earth’s surface where R is earth’s radius. The time period of the pendulum will be (CPMT 89)

(a) Zero (b)

(c) 4 s (d) Infinite

10. The pendulum is acts as second pendulum on earth. Its time on a

planet, whose mass and diameter are twice that of earth, is (MHT-CET-2005)

(a) (b) 2 s

(c) (d) 1/

11. A particle of mass m is hanging vertically by an ideal spring of

force constant K. if the mass is made to oscillate vertically, its total energy is (CPMT 78)

(a) Maximum at extreme position

(b) Maximum at mean position

(c) Minimum at mean position

(d) Same at all positions

12. At a place where g 980 cm/sec2. the length of seconds

pendulum is about

(a) 50 cm (b) 100 cm

(c) 2 cm (d) 2 m 13. The maximum velocity for particle in SHM is 0.16 m/s and

maximum acceleration is 0.64 m/s2. The amplitude is (MHT-CET-2004)

(a) 4 102 m (b) 4 101 m

(c) 4 10 m (d) 4 100 m

14. A particle is vibrating in S.H.M. with an amplitude of 4 cm. at

what displacement from the equilibrium position is its energy half potential and half kinetic?(NCERT 84)

(a) 2.5 cm (b)

(c) 3 cm (d) 2 cm

15. The time period of a spring pendulum is (CPMT 71)

(a) (b)

(c) (d)

16. The equation of displacement of particle performing SHM is X

= 0.25 sin (200 t). The maximum velocity is (MHT-CET-2004)

(a) 100 m/s (b) 200 m/s

(c) 50 m/s (d) 150 m/s

17. A pendulum suspended from the roof of a train has a period T

(When the train is at rest). When the train is accelerating with a uniform acceleration ‘a’, the time period of the pendulum will (NCERT 80)

(a) Increase (b) Decrease

(c) Remain unaffected (d) Become infinite

18. A particle executing a vibratory motion while passing through

the mean position has (CPMT 92)

(a) Maximum P.E. and minimum K.E.

(b) Maximum K.E. and minimum P.E.

(c) P.E. and K.E. both maximum

(d) P.E. and K.E. both minimum

19. The frequency of wave is 0.002 Hz. Its time period is (MHT-

CET-2004)

(a) 100 s (b) 500 s

(c) 5000 s (d) 50 s

20. A simple pendulum has a period T. it is taken inside a lift

moving up with uniform acceleration g/3. now its time period will be (NCERT 90)

(a) (b)

(c) (d) 21. For a magnet of time period T magnetic moment is M, if the

magnetic moment becomes one fourth of the initial value, then the time period of oscillation becomes.

(MHT CET 2006)

(a) Half of initial value

(b) One fourth of initial value

(c) Double of initial value

(d) Four time initial value

22. The value of displacement of particle performing SHM, when

kinetic energy is (3/4)th of its total energy is (MHT-CET-2004)

(a) (b)

(c) (d)

23. The shape of l T graph of simple pendulum is,

(CPMT-92)

(a) Curve (b) Parabola

(c) Straight line (d) Hyperbola

24. A simple pendulum is suspended from the roof of a trolley

which moves in a horizontal direction with an acceleration ‘a’ then the time period is given

T 2 , where g is equal to (CBSE 91)

(a) ag (b) 3 a

(c) g a (d)

25. Two equal negative charges q are fixed at point (0, a) and (0,

a) on the Y-axis A positive charge q is released from rest at point (2a, 0) on the X-axis. The charge Q will (IIT 83)

(a) Execute simple harmonic motion about the origin

(b) Move to the origin and remained at rest

(c) Move to infinity

(d) Execute oscillatory motion but not simple harmonic motion

1. Answer: (d)

2. Answer: (b)

3. Answer: (a)

4. Answer: (b)

5. Answer: (d)

6. Answer: (c)

7. Answer: (c)

8. Answer: (b)

9. Answer: (d)

10. Answer: (c)

11. Answer: (d)

12. Answer: (b)

13. Answer: (a)

14. Answer: (d)

15. Answer: (a)

17. Answer: (b)

18. Answer: (b)

19. Answer: (b)

20. Answer: (c)

21. Answer: (c)

22. Answer: (a)

23. Answer: (b)

24. Answer: (d)

25. Answer: (d)

1. The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the mass of the particle is 0.32 kg, then what is the maximum velocity of the particle? (MHT-CET-2004)

(a) 5 m/s (b) 15 m/s

(c) 10 m/s (d) 20 m/s

2. If a hole is bored along the diameter of the earth and a stone is

dropped into the hole (CPMT 84)

(a) The stone reaches the centre of the earth and stops there

(b) The stone reaches the other side of the earth and stops there

(c) The stone executes simple harmonic motion about the centre of the earth

(d) The stone reaches the other side of the earth and escapes into space

3. A simple pendulum of length L and mass m is oscillating in a

plane about a vertical line between angular limits - and. For an angular displacement

The tension in the string and the velocity of the bob are T and respectively. The following relation holds good under the above conditions (IIT 86)

(a) T cos mg

(b) T m g cos

(c) T mg cos

(d) Tangential acceleration aT g sin

4. When a particle performing uniform circular motion of radius

10 cm undergoes the SHM, what will be its amplitude? (MHT-CET-2004)

(a) 10 cm (b) 5 cm

(c) 2.5 cm (d) 20 cm

5. The work done by the tension in the string of a pendulum

during one complete vibration is equal to

(NCERT 83)

(a) Potential energy of pendulum

(b) Total energy of pendulum

(c) Kinetic energy of pendulum

(d) Zero

6. A particle is executing simple harmonic motion with an

amplitude a. when its kinetic energy is equal to its potential energy its distance from the mean position

(CPMT 90, PMT MP 87)

(a) (b)

(c) (d)

7. A magnet of magnetic M oscillates in magnetic field B with

time period 2 sec. If now the magnet is cut into two half pieces parallel to the axis, then what is new time period if only one part oscillate in field? (MHT-CET-2004)

(a) 2 s (b) 2 s

(c) (d) 2.4 s

8. Time period of simple pendulum of length l and a place where

acceleration due to gravity is g is T. what is the period of a simple pendulum of the same length at a place where the acceleration due to gravity is 1.029 is,

(CPMT 82)

(a) T (b) 1.02 T

(c) 0.99 T (d) 1.01 T

9. The potential energy of a particle with displacement X is U

(X). the motion is simple harmonic, when

(K is a positive constant) (CPMT 82)

(a) (b) U KX2

(c) U K (d) U KX

10. A particle is subjected to two S.H.M.s x1 A1 sin t and x2

A2 sin The resultant S.H.M. will have an amplitude of (IIT 96)

(a) (b)

(c) (d) A1 A2

11. If velocity of a body is half the maximum velocity. Then what

is the distance from the mean position?

(MHT CET 2002, 2003)

(a) 2a (b)

(c) a (d)

12. If the length of the simple pendulum is increased by 44%,

then what is the change in time period of pendulum? (MHT-CET-2004)

(a) 22% (b) 20%

(c) 33% (d) 44% 13. A linear harmonic oscillator of force constant 2 106 N/m and

amplitude 0.01 m has a total mechanical energy of 100 J. it’s maximum potential energy is

(IIT 89)

(a) 100 J (b) 200 J

(c) 150 J (d) 0

14. The period of oscillation of a simple pendulum of constant

length at earths surface is T, it period inside a mine is (CPMT 73)

(a) Greater than T.

(b) Less than T.

(c) Equal to T.

(d) Cannot be compared

15. A spring having a spring constant k is loaded with a mass m. the

spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is (NCERT 90)

(a) (b) k

(c) 2k (d) k2

16. Dimensions of force constant are (MHT CET 2003)

(a) [M1 L0 T2] (b) [M1 L0 T2]

(c) [M1 L1 T2] (d) [M1 L1 T2]

17. Spring is pulled down by 2 cm. What is amplitude of motion?

(MHT-CET-2003)

(a) 0 cm (b) 6 cm

(c) 2 cm (d) 4 cm

18. The particle is performing S.H.M. along a straight line with amplitude ‘a’. the potential energy is maximum when the displacement is (CPMT 82)

(a) A (b)

(c) 0 (d)

19. The period of a simple pendulum is doubled when

(CPMT 74)

(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.

20. Particle moves from extreme position to mean position, its

(MHT-CET-2003)

(a) Kinetic energy increases, potential increases decreases

(b) Kinetic energy decreases, potential increases

(c) Both remains constant

(d) Potential energy becomes zero and kinetic energy remains constant

21. If a particle is moving in a circle, with a uniform speed, then its motion is, (CPMT 78)

(a) Oscillatory (b) Periodic

(c) Non-periodic (d) Simple harmonic

22. A simple pendulum performs simple harmonic motion about x

0 with an amplitude A and time period T. the speed of the

pendulum at will be (PMT-MP 87)

(a) (b)

(c) (d)

23. If velocity of body is half the maximum velocity. Then what is

the distance from the mean position? (MHT-CET-2003)

(a) 2 A (b)

(c) A (d)

24. The necessary and sufficient condition for S.H.M. is

(NCERT 74)

(a) Constant period

(b) Constant acceleration

(c) Proportionality between restoring force and displacement from equilibrium position in opposite direction

(d) None of the above

25. The motion of a particle executing simple harmonic motion is

given by X 0.01 sin 100 (t 0.05), where X is in metres ant t in second. The time period in second is (CPMT 90)

(a) 0.001 (b) 0.02

(c) 0.1 (d) 0.2

1. Answer: (c)

2. Answer: (c)

3. Answer: (b)

4. Answer: (a)

5. Answer: (d)

6. Answer: (b)

7. Answer: (a)

8. Answer: (c)

9. Answer: (b)

10. Answer: (c)

11. Answer: (b)

12. Answer: (b)

13. Answer: (a)

14. Answer: (a)

15. Answer: (c)

16. Answer: (a)

17. Answer: (c)

18. Answer: (a)

19. Answer: (c)

20. Answer: (a)

21. Answer: (b)

22. Answer: (c)

23. Answer: (b)

24. Answer: (c)

25. Answer: (b)