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Q1. If the length of spring is halved, then its force constant becomes

• Halved
• Double
• Thrice
• Remains same

Solution

The spring constant/force constant is inversely proportional to the length of the spring. Hence, if the length is halved, the force constant will be doubled.

Q2. Equation of a body in S.H.M is x = 5 cos (3πt + π/3). Its amplitude and velocity at t = 2 s would be

• 40.97 m/s
• -41.5 m/s
• -40.79 m/s
• 40.5 m/s

Solution

Q3. Amplitude of an oscillator reduces to one fourth of its initial value when it completes 200 oscillations. What will be its amplitude, when it completes 400 oscillations.

• 1/16
• 1/10
• 4/16
•   ½

Solution

Q4. A pendulum clock that work fines on earth is placed on the moon. How many times the clock would tick in an actual time of 30 seconds? Given object weighs one sixth on moon as much as on earth.

• 12.25
• 12.0
• 11.5
• 13

Solution

 

Q5. Under forced oscillation, the phase of harmonic motion of the particle:

• is equal to the phase of the driving force.
• is greater than the phase of the driving force
• is less than the phase of the driving force
• differs from the phase of the driving force

Solution

Under forced oscillation, the phase of harmonic motion of the particle differs from the phase of the driving force.

Q6. The motion of simple pendulum is said to be S.H.M when its angle θ through which bob is displaced from its equilibrium position is

• θ is very large  
• θ is zero  
• θ is very small  
• θ = cosθ  

Solution

The motion of simple pendulum is said to be S.H.M when its angle θ through which bob is displaced from its equilibrium position is very small

Q7. Under forced oscillation, the phase of the harmonic motion of the particle and phase of driving force

• Are same
• Are different
• Both are zero
• Not present

Solution

Under forced oscillation, the phase of the harmonic motion of the particle and phase of driving force are different.

Q8. A body following S.H.M on straight line path has an amplitude of oscillations = 4 cm. The magnitude of its acceleration is equal to that of its velocity when body is 2 cm away from its mean position. Find the time period.

• 2 s
• 1.8 s
• 1.6 s
• 2.1 s

Solution

Q9. If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is

• o.1 m/s
• 0.15 m/s
• 0.8 m/s
• 0.16 m/s

Solution

Q10. A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is

• 5 N
• 4 N
• 0.5 N
• 0.15 N

Solution

The basic expression for time period is . Hence, we get . We know that k is the force for unit displacement. Hence, the maximum force is k time maximum displacement.

Q11. Value of spring constant depends upon

• Elastic properties of the spring
• Length of spring
• Number of turns
• Mass attached to spring

Solution

Value of spring constant depend on the elastic properties of the spring.

Q12. A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately:

• y = .10 sin 9.9t
•  y = .10 cos 9.9t
• y = .10 cos (9.9t + .1)
• y = .05 cos 9.9t

Solution

Its position as a function of time is approximately y = .05 cos 9.9t

Q13. Due to what force a simple pendulum remains in simple harmonic motion?

• mg sinθ, component of weight due to gravitational force
• mg , weight
• a, acceleration
• y, displacement

Solution

Mg sinθ, component of weight due to gravitational force.

Q14. In simple harmonic motion, motion is executed by a particle that is subject to a force which is ___________to the displacement of the particle and is directed towards the______________.

• proportional, extreme position
• inversely proportional, mean position
• proportional, mean position
• inversely proportional, extreme position

Solution

In simple harmonic motion, motion is executed by a particle that is subject to a force which is proportional to the displacement of the particle and is directed towards the mean position.

Q15. Particle with amplitude A, displacement Y and time period T starts in SHM from its mean position. What would be particle's displacement when its speed is half of the maximum speed?

• 3A/5
• √3 A/2  
• √2 A/3
• A/√3

Solution

Q16. If the sign in equation F=-kx is changed what would happen to the motion of the oscillating body?

• Motion would be linearly accelerated motion
• Body would come at rest  
• Motion would be oscillating accelerated
• Body would slow down

Solution

If the sign is changed in F = -kx, then the force, and hence, the acceleration will not be opposite to the displacement. Due to this the particle will not oscillate and would accelerate in the direction of displacement. Therefore, the motion of the body will become linearly accelerated motion.

Q17. A spring compressed by 0.2 m develops a restoring force 10 N. A body of mass 5 Kg is placed on it. What would be the force constant k of the spring and the depression of the spring y under the weight of the body?

• 40 Nm^-1, 0.5m
• 50 Nm^-1,0.5m
• 0.5 Nm^-1,0.5m
• 60 Nm^-1,0.5m

Solution

Q18. What determines the natural frequency of a body?

• Mass and speed of the body
• Elastic properties and dimensions of the body
• Position of the body with respect to force applied
• Oscillations in the body

Solution

The natural frequency of a body is determined by elastic properties and dimensions of the body.

Q19. At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?

• a√2
• 2√a
• a
• a/2

Solution

Q20. The amplitude 'a' of a simple harmonic oscillator (with period, T and energy, E) is tripled. What would happen to T and E?

• T and E triples
• T gets tripled and E becomes zero
• T remains same and E becomes nine times
• T and E doubles

Solution

Q21. When a particle moves in a simple harmonic motion, its acceleration at the ends of its path is:

• zero
• less than at equilibrium
• less than g
• maximum

Solution

When a particle moves in a simple harmonic motion, its acceleration at the ends of its path is zero.

Q22. The amplitude of a simple harmonic oscillator is doubled; then period of oscillator would

• Double
• Remain same
• Four times
• Become half

Solution

As period of oscillator depends upon inertia factor and spring factor which in turn are independent of amplitude. So, the period of simple harmonic oscillator would remain the same on doubling the amplitude.

Q23. If the strong external periodic driving force matches one or more of the natural frequencies of the mechanical structure like bridge, aero plane, buildings then the resulting oscillations could

• Move the structure
• Cause echo in structure
• Rupture /damage the structure
• Would have no effect on structure

Solution

If the strong external periodic driving force matches one or more of the natural frequencies of the mechanical structure like bridge, aero plane, buildings then the resulting oscillations could rupture or damage the structure.

Q24. If the length of a second's pendulum is increased thrice, what would be its time period?

• 2.35s
• 6.02s
• 3.46s
• 2.08s

Solution

Q25. A mass of 0.3 Kg is attached to the end of a spring. If another mass of 0.04 Kg is added to the end of the spring, it stretches by 12 cm more. What would be the period of vibration of the system if 0.04 kg mass is removed?

• 2.1 s
• 1.7 s
• 2.2 s
• 1.9 s

Solution

Q26. A hollow sphere filled with water and one small hole at bottom is hung by a long thread and made to oscillates.  What would be the effect on the period of oscillations as water slowly flows out of the hole at bottom?

• Keep on increasing
• First increases then decreases
• First decreases then increases
• Keeps on decreasing

Solution

As water slowly leaks out, the centre of gravity of water and sphere shifts down from the centre of the sphere. Due to which the length of the simple pendulum increases. This continues till half of the sphere is emptied. After that the length of the simple pendulum decreases. Hence, the time period initially increases and then decreases since it is directly proportional to length.

Q27. The restoring force in a simple harmonic motion is _________ in magnitude when the particle is instantaneously at rest.

• zero
• maximum
• minimun
• Not defined

Solution

The restoring force in a simple harmonic motion is maximum in magnitude when the particle is instantaneously at rest.

Q28. If we triple the mass on a simple pendulum, the frequency will be__________

• Three times what it was.
• Six times what it was.
• Remain same
• One-third what it was.

Solution

The frequency of pendulum does not depend on the mass. Hence, it remains unchanged.

Q29. The work done by the string of a simple pendulum in S.H.M is

• mg
• equal to kinetic energy of the system
• zero
• equal to total energy of system

Solution

The work done by the string of a simple pendulum in S.H.M is zero.  

Q30. Damping force on a system executing S.H.M

• Zero
• Constant
• Varies with speed of the system
• Twice

Solution

Damping force on a system executing S.H.M varies with the speed of the system.

Q31. The equation of motion of a particle is x = 3 cos(0.45 t + π/4) m. Its maximum acceleration is

• 0.50 ms^-2
• 0.60 ms^-2
• 0.55 ms^-2
• 0.45 ms^-2

Solution

Maximum acceleration is given as a_max = ω²a = (0.45)² x 3 = 0.60 ms^-2

Q32. If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes

• circular motion
• motion along x axis
• motion along y axis
• Simple Harmonic motion

Solution

Simple Harmonic motion

Q33. Find the amplitude of the S.H.M whose displacement y in cm is given by equation y = 3 sin157t + 4cos157t where t is time in seconds.

• 50Hz
• 20Hz
• 40Hz
• 25Hz

Solution

Q34. The reason why oscillations becomes damped is _________

• Normal force
• Tangential force
• Friction
• Equidistant force

Solution

Friction is the reason why oscillations becomes damped 

Q35. Time period of a simple pendulum clock at hills or in mines would

• Increase
• Slow down
• Become zero
• Remain same

Solution

Time period, T is proportional to 1/√g therefore value of T will increase with decrease in the value of g at hills or in mines. Due to this the pendulum clock will slow down.

Q36. Epoch is measured in

• Radians
• Meters
• gram/cm
• seconds

Solution

Epoch is measured in Radians.

Q37. What provides the restoring force for SHM in case of column of mercury in U-tube?

• Temperature
• Elasticity
• Height of column
• Weight of mercury

Solution

Weight of mercury provides the restoring force for SHM in case of column of mercury in U-tube.

Q38. Potential energy of a particle with mass m is U = kx³, where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period 'T' is

• Proportional to √a
• Independent  of a
• Proportional to a²
• Proportional to 1/√a

Solution

Q39. A second pendulum is mounted in a space shuttle. Its period of oscillation will decrease when rocket is

• ascending up with uniform acceleration
• descending down with uniform acceleration
• moving in geostationary orbit
• moving up with uniform velocity

Solution

Time period decreases when effective value of g increases, this happens when rocket accelerates upwards.

Q40. The amplitude of S.H.M at resonance is _______ in the ideal case of zero damping.

• Maximum
• Zero
• Minimum
• Infinite