Oscillations Easy Questions and Answers | Page - 2

Oscillations

A particle moving along the X-axis, executes simple harmonic motion then the force acting on it is given by

1. – A kx
1. A cos (kx)
1. A exp (– kx)
1. Akx
  where, A and k are positive constants.

View Hint View Answer Discuss in Forum

For simple harmonic motion, F = – Kx.
Here, K = Ak.

Correct Option: A

For simple harmonic motion, F = – Kx.
Here, K = Ak.

The particle executing simple harmonic motion has a kinetic energy K₀ cos² ω t. The maximum values of the potential energy and the total energy are respectively

1. K₀ / 2 and K₀
1. K₀ and 2K₀
1. K₀ and K₀
1. 0 and 2K₀.

View Hint View Answer Discuss in Forum

We have, U + K = E
where, U = potential energy, K = Kinetic energy, E = Total energy.
Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and vice-versa.
∴ U_max - 0 = E ⇒ U_max = E
Further,
K.E. = 1 mω² a² cos² ωt
2

But by question , K.E. = K₀ cos² ωt
K₀ = 1 mω² a²
2

Hence, total energy, E = 1 mω² a² = K₀
2

∴ U_max = K₀ & E = K₀

Correct Option: C

We have, U + K = E
where, U = potential energy, K = Kinetic energy, E = Total energy.
Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and vice-versa.
∴ U_max - 0 = E ⇒ U_max = E
Further,
K.E. = 1 mω² a² cos² ωt
2

But by question , K.E. = K₀ cos² ωt
K₀ = 1 mω² a²
2

Hence, total energy, E = 1 mω² a² = K₀
2

∴ U_max = K₀ & E = K₀

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is

1. 8 U
1. 16 U
1. U / 4
1. 4 U

View Hint View Answer Discuss in Forum

The potential energy of a spring = 1 kx²
2

U = 1 k.(2)² = 4 × 1 k
2 2

For x = 8 cm,
Energy stored = 1 k.(8)² = 64 × 1 k
2 2

= 64 × U = 16 U
4

Correct Option: B

The potential energy of a spring = 1 kx²
2

U = 1 k.(2)² = 4 × 1 k
2 2

For x = 8 cm,
Energy stored = 1 k.(8)² = 64 × 1 k
2 2

= 64 × U = 16 U
4

A linear harmonic oscillator of force constant 2 × 10⁶ N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Its

1. potential energy is 160 J
1. potential energy is 100 J
1. potential energy is zero
1. potential energy is 120 J

View Hint View Answer Discuss in Forum

Force constant k = 2 × 10⁶ N/m
Amplitude (x) = 0.01 m
Potential Energy = 1 kx²
2

= 1 × (2 × 10⁶) × (0.01)² = 100 J
2

Correct Option: B

Force constant k = 2 × 10⁶ N/m
Amplitude (x) = 0.01 m
Potential Energy = 1 kx²
2

= 1 × (2 × 10⁶) × (0.01)² = 100 J
2

In a simple harmonic motion, when the displacement is one-half the amplitude, what fraction of the total energy is kinetic?

1. 0
1. 1
  4
1. 1
  2
1. 3
  4

View Hint View Answer Discuss in Forum

Total energy of particle executing S.H.M. of amplitude (A).
E = 1 mω² A²
2

K.E.of the particle = 1 mω² A² - A² when x = A
2 4 2

= 1 mω² × 3 A² = 1 × 3 mω² A²
2 4 2 4

Clearly , KE = 3
Total energy 4

Correct Option: D

Total energy of particle executing S.H.M. of amplitude (A).
E = 1 mω² A²
2

K.E.of the particle = 1 mω² A² - A² when x = A
2 4 2

= 1 mω² × 3 A² = 1 × 3 mω² A²
2 4 2 4

Clearly , KE = 3
Total energy 4