Question 1

Which of the following conditions is essential for a motion to be Simple Harmonic Motion (SHM)?

Accepted Answer

In Simple Harmonic Motion (SHM), the restoring force or acceleration is directly proportional to the displacement from the mean position and acts in the opposite direction to the displacement ($a = -\omega^2 x$).

Question 2

The phase difference between displacement and velocity of a particle executing Simple Harmonic Motion is:

Accepted Answer

If displacement $x = A \sin(\omega t)$, then velocity $v = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin\left(\omega t + \frac{\pi}{2}\right)$. Thus, velocity leads displacement in phase by $\pi/2$ radians or $90^\circ$.

Question 3

The time period of a seconds pendulum is:

Accepted Answer

A seconds pendulum is defined as a pendulum whose time period is exactly 2 seconds. It takes 1 second to move from one extreme to the other.

Question 4

If the length of a simple pendulum is increased by 4 times, its time period becomes:

Accepted Answer

The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$. Since $T \propto \sqrt{l}$, if $l$ becomes $4l$, new time period $T' = \sqrt{4} T = 2T$.

Question 5

The maximum velocity of a particle executing SHM with amplitude $A$ and angular frequency $\omega$ is given by:

Accepted Answer

Velocity in SHM is $v = \omega \sqrt{A^2 - x^2}$. Maximum velocity occurs at the mean position ($x = 0$), giving $v{\max} = A\omega$.

Oscillations — NCERT Class 11 Physics Important Questions & Revision Notes