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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

For a simple pendulum why is a spherical bob chosen.

Answer»

A spherical bob is chosen for a simple pendulum for the following reasons :

  1. For a given value of density, the spherical shape has minimum volume.
  2. The air friction is minimum due to spherical shape.
2.

What is the relation between oscillating frequency and total energy of a simple pendulum?

Answer»

For a simple pendulum
Etotal = \(\frac { 1 }{ 2 } \) mω2 a2 = \(\frac { 1 }{ 2 } \)m(2πn)2 a2
\(\frac { 1 }{ 2 } \) m 4π2 n2 a2
= 2π2 mn2 a2
Here, Etotal = total energy
m = mass of the pendulum (bob)
ω = angular frequency = 2πn = 2π/T
n = linear frequency
a = amplitude.

3.

In simple harmonic motion what is the phase difference between displacement and acceleration?

Answer»

The phase difference between displacement and acceleration is π or 180°.

4.

What will be the initial phase angle of a body starting its simple harmonic motion from the extreme end?

Answer»

The time taken by the body starting its simple harmonic motion from the extreme end to the mean position is T/4 and accordingly the phase angle is π/2. Hence, the initial phase angle is π/2.

5.

In simple harmonic motion which physical quantity is conserved?

Answer»

Total mechanical energy is conserved in simple harmonic motion.

6.

In simple harmonic motion the value of displacement is how much phase angles ahead of its acceleration value?

Answer»

In a simple harmonic motion, the displacement and acceleration are in the opposite direction, displacement is ahead of acceleration by a phase angle π.

7.

For a simple harmonic motion what is the relationship between acceleration and displacement?

Answer»

For a simple harmonic motion, acceleration is directly proportional to the displacement of the object from its mean position and always points towards the mean position (f = -ω2y).

8.

What will be the effect on time period on using soft spring of same length instead of hard spring?

Answer»

The relation between the time period and the spring constant of an oscillating body is

\(T = 2\pi \sqrt{\frac{m}{k}}\)

k = spring constant
The spring constant for a softer spring is less than that for a harder one.
i.e., Ksoft < Khard
Hence from the above equation
Tsoft > Thard
i. e, the time period will increase.

9.

What will be the value of the time period of a simple pendulum if the amplitude is reduced to half of its original value?

Answer»

Formula for the time period of a simple pendulum,

\(T = 2 \pi \sqrt{\frac{l}{g}}\)

In this expression, there is no term ‘a’, hence the time period is independent of the amplitude of oscillation. Thus, the time period remains unaffected.

10.

What do you understand by damped oscillations?

Answer»

If there is only restoring force acting on a vibrating object for example simple pendulum or tuning fork then that object can vibrate till infinity. This type of vibration or oscillation is called ‘free oscillation’ and this type of vibration is called ‘free vibration’. Its frequency is called ‘Natural Frequency’. There is only restoring force working in free vibration. Its amplitude and energy both are constant with time.