Oscillation and Waves MCQs

Ans. d) \(\frac{T}{12}\) sec

Explanation: For the given case the equation of motion be x= Asinωt
At t = 0, x = 0; so, the Particle is at its mean position.
The particle will be at x =\(\frac{A}{2}\) and time will be t,
Where \(\frac{A}{2}\) = Asinωt 

Or sinωt = \(\frac{1}{2}\)

Or ωt = \(\frac{\pi}{2}\)
(Minimum value of wt is taken because we will calculate the time taken by particle x = 0 to directly to x =) \(\frac{A}{2}\)
Thus, t = \(\frac{\pi}{6\omega}\)

= \(\frac{\pi}{6(2 \pi) T}\)

=\(\frac{T}{12}\) sec.

Ans. (c) A²

Explanation: The equation for energy of particle in SHM is given by-

E= \(\frac{1}{2}\) Ka ²

• E = Energy of particle exhibiting simple harmonic motion
• K = Spring constant
• A = maximum amplitude of motion.

Thus, E is directly proportional to the square of amplitude in Simple Harmonic Motion.

Ans. (b) It is proportional to position linearly

Explanation: In Simple Harmonic Motion, the force on the oscillating body is always directed toward it’s mean position. When x = A sin (ωt)

 F = - m A ω ² sin (ωt). 

Thus, the force is proportional to position in a linear fashion.

Ans. ( c) ; m = k/ω²

Explanation: The force on a particle exhibiting Simple Harmonic Motion is F = – m ω ² x (t) (Equation-I)

Here, F = – kx (t) (Equation II). 

Equating (I) and (II) :-

→ -m ω² x = – kx

→ m = k/ω² or k =m ω²

So, the relationship in terms of angular frequency is m = k/ω² or k =m ω²

Ans. b) π√2 s

Explanation: F = -m ω 2 (x) 

→ 2 = 1ω² ;

→ω = √2

As we know that k=mω² 

Time period = T = 2π/ω

T = π√2 s.

As the time period is π√2 seconds. So option b) is correct.

Ans. (b) The direction of velocity is towards the amplitude, whereas the direction of acceleration is towards the mean position.

Explanation: In Simple Harmonic Motion, the force on the oscillating particle is always directed towards the mean position (exception is when the particle is in the mean position). Factually, the particle is currently approaching to an extreme position, the direction of velocity will be in the direction of the amplitude.

Ans. (a) 1:8 

Explanation: The question clearly states that the displacement of the body (in SHM) is equal to one-third of it’s amplitude, so it implies that:

x=1/3 a

Kinetic energy = \(\frac{1}{2}\) mω²(A² - x²)

K.E. = \(\frac{1}{2}\) mω² (a² – \(\frac{1}{3}\)a)

= \(\frac{8}{9}\) × \(\frac{1}{2}\) m ω 2a ² (1)

= Potential energy = \(\frac{1}{2}\)mω²( \(\frac{1}{3}\)A)² = \(\frac{1}{9}\) ×  \(\frac{1}{2}\) mω²A² --- (2)

From equation 1 and 2, the ratio of potential energy to kinetic energy is =

P_E : K _E = \(\frac{ \frac{1}{9} × \frac{1}{2} m \omega 2 A 2}{\frac{8}{9} × \frac{1}{2} m \omega 2 a 2} = \frac{1}{8}\)

Thus, the option a) is correct.

Ans. (a) bob

Explanation: A simple pendulum is one which has a piece of a small sized stone or a metal ball suspended from a rigid support by a thin inextensible thread. Every simple pendulum consists of the following three parts:

• cable or wire
• bob or weight,
• fixed point

Ans. (d) Infinite

Reason. Thetime period of the pendulum: T= 2 π √L/g

where, L = length of string, g = acceleration due to gravity (m/s2)

as T ∝ 1/g

The net gravity inside an artificial satellite in the earth’s orbit is zero. This fulfils the condition for the time period to be infinite.

T= 2 π √L/g 

= T= 2 π √L/0 =

T= ∞

Hence, proved that the time period is infinite.

Ans. (b) Increase

Explanation: The time period of the pendulum: T= 2 π √L/g

where, L = length of string, g = acceleration due to gravity (m/s2)

as T ∝ √L

From the above calculation we can infer that the time period is inversely proportional to the length of the string of the pendulum. Thus, the time period increases as the length increases.