Question

A particle is initially at the centre and going towards the left. Let $ T $ be the time period of the SHM it is undergoing. What will be its position and velocity at time $ 3T/4 $, if it starts from the centre at $ t = 0 $?

• A. At right extreme, zero velocity
• B. At centre, maximum speed towards left
• C. At centre, maximum speed towards right
• D. Mid-way between centre and -A

Hint:

In SHM, the particle's velocity is maximum when it passes through the centre of motion, and its direction of motion depends on the phase at that instant.

Answer 1

The Correct Option is B

Step 1: Understand the concept of SHM.
In Simple Harmonic Motion (SHM), the displacement of the particle is given by: \[ x = A \cos(\omega t + \phi) \] where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is the time, and \( \phi \) is the phase constant.
Step 2: Given conditions.
At \( t = 0 \), the particle is at the centre, so \( x(0) = 0 \).
The particle is initially moving towards the left, which means the velocity is negative.

Step 3: Velocity in SHM.
The velocity in SHM is given by: \[ v = -A \omega \sin(\omega t + \phi) \] Since the particle starts at the centre and moves to the left, the phase constant \( \phi \) is \( 0 \).
So, the displacement and velocity equations become: \[ x = A \cos(\omega t) \quad \text{and} \quad v = -A \omega \sin(\omega t) \]
Step 4: Find the position and velocity at \( t = 3T/4 \). At \( t = 3T/4 \), the angular position is: \[ x = A \cos\left(\omega \cdot \frac{3T}{4}\right) = A \cos\left(\frac{3\pi}{2}\right) = 0 \] So, the particle is at the centre.
Step 5: Find the velocity. The velocity at \( t = 3T/4 \) is: \[ v = -A \omega \sin\left(\frac{3\pi}{2}\right) = A \omega \] Thus, the particle is moving with maximum speed towards the left.