Oscillations NEET MCQ

1. Which of the following quantities remains constant during simple harmonic motion (SHM)?

a) Displacement
b) Velocity
c) Acceleration
d) Total energy

Answer:

d) Total energy

Explanation:

In SHM, although kinetic and potential energies change, their sum, which is the total mechanical energy, remains constant.

2. In a simple pendulum, at what position is the potential energy maximum?

a) Mean position
b) Extreme position
c) Halfway between mean and extreme position
d) None of the above

Answer:

b) Extreme position

Explanation:

At the extreme position, the pendulum has reached its maximum displacement, and thus, its potential energy is at a maximum.

3. If the amplitude of a simple harmonic oscillator is doubled, its period will:

a) Remain unchanged
b) Double
c) Halve
d) Quadruple

Answer:

a) Remain unchanged

Explanation:

The period of SHM does not depend on amplitude.

4. What kind of force acts on a body performing SHM?

a) Constant force
b) Variable force
c) No force
d) Centripetal force

Answer:

b) Variable force

Explanation:

A restoring force proportional to its displacement and acting in the opposite direction is always acting on a body in SHM.

5. The time period of a simple pendulum of length L on the surface of Earth is T. What will be its time period on a planet where acceleration due to gravity is half of that on Earth?

a) T
b) T/√2
c) √2 T
d) 2 T

Answer:

c) √2 T

Explanation:

Time period T is directly proportional to 1/√g. If g is halved, T will become √2 times.

6. For a spring-block oscillator, the time period T is given by:

a) T = 2π√(m/k)
b) T = 2π√(k/m)
c) T = π√(m/k)
d) T = π√(k/m)

Answer:

a) T = 2π√(m/k)

Explanation:

The time period for a spring-block system is given by T = 2π√(m/k), where m is the mass and k is the spring constant.

7. Which of the following is a necessary condition for SHM?

a) Force must be maximum at mean position
b) Force must be minimum at mean position
c) Force must be zero at extreme position
d) Force must be zero at mean position

Answer:

d) Force must be zero at mean position

Explanation:

For SHM, the restoring force is zero at the mean position.

8. The displacement of a body performing SHM is given by x = A sin(ωt + φ). Here, φ represents:

a) Frequency
b) Angular frequency
c) Amplitude
d) Initial phase

Answer:

d) Initial phase

Explanation:

In the equation, φ represents the initial phase or phase constant.

9. The maximum velocity of a body in SHM is given by:

a) ωA
b) ω/T
c) A/T
d) A/ω

Answer:

a) ωA

Explanation:

The maximum velocity Vmax = ωA, where ω is the angular frequency and A is the amplitude.

10. In SHM, acceleration is directly proportional to:

a) Displacement
b) Velocity
c) Time period
d) Frequency

Answer:

a) Displacement

Explanation:

Acceleration in SHM is directly proportional to the displacement and is directed towards the mean position.

11. The potential energy of a harmonic oscillator is zero at:

a) Mean position
b) Extreme position
c) Both mean and extreme positions
d) None of the above

Answer:

a) Mean position

Explanation:

At the mean position, the displacement is zero; hence, potential energy is zero.

12. If the frequency of a harmonic oscillator is 10 Hz, its time period will be:

a) 0.01 s
b) 0.1 s
c) 1 s
d) 10 s

Answer:

b) 0.1 s

Explanation:

Time period T = 1/frequency. Therefore, T = 1/10 = 0.1 s.

13. A body is in SHM. When its displacement is half of the amplitude, what fraction of its total energy is kinetic?

a) 1/4
b) 1/2
c) 3/4
d) 7/8

Answer:

c) 3/4

Explanation:

When displacement is half of the amplitude, the kinetic energy is 3/4th of the total energy.

14. The angular frequency (ω) of a body in SHM is related to the time period (T) as:

a) ω = 2πT
b) ω = T/2π
c) ω = 2π/T
d) ω = πT/2

Answer:

c) ω = 2π/T

Explanation:

Angular frequency ω = 2π times the frequency, and frequency = 1/T. Thus, ω = 2π/T.

15. A body is oscillating with a time period of 2 seconds. Its frequency is:

a) 0.5 Hz
b) 2 Hz
c) 1 Hz
d) 4 Hz

Answer:

a) 0.5 Hz

Explanation:

Frequency = 1/Time period. Therefore, frequency = 1/2 = 0.5 Hz.

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