Q14.24(c):

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (c) 0 cm.

Step 1: Given Variables & Initial Setup

To analyze the Simple Harmonic Motion (SHM) of the body, we extract the primary parameters provided:

• Amplitude ($A$): $5\text{ cm} = 0.05\text{ m}$
• Period ($T$): $0.2\text{ s}$
• Displacement ($x$): $0\text{ cm} = 0\text{ m}$

We calculate the angular frequency ($\omega$) using the relationship between period and frequency [Per the definition of SHM]:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\text{ rad/s}$

Step 2: Determining Velocity at Displacement $x=0$

The velocity ($v$) of a body in SHM is defined by the function [Derived from the differentiation of $x(t) = A \sin(\omega t + \phi)$]:

$v = \omega \sqrt{A^2 - x^2}$

Substituting $x = 0$ into the equation:

$v = (10\pi) \sqrt{(0.05)^2 - 0^2}$

$v = (10\pi) \times 0.05$

$v = 0.5\pi\text{ m/s} \approx 1.57\text{ m/s}$

Step 3: Determining Acceleration at Displacement $x=0$

The acceleration ($a$) of a body in SHM is defined by the restoring force relationship [Per Newton’s Second Law and Hooke’s Law]:

$a = -\omega^2 x$

Substituting $x = 0$ into the equation:

$a = -(10\pi)^2 \times 0$

$a = 0\text{ m/s}^2$

Visual Representation of SHM Parameters

Summary Analysis

At the equilibrium position ($x=0$), the body possesses maximum potential energy transformation into kinetic energy, resulting in maximum velocity. Conversely, since the restoring force $F = -kx$ is directly proportional to displacement, the acceleration vanishes at the equilibrium point.

Final Solution: The velocity is $1.57\text{ m/s}$ and the acceleration is $0\text{ m/s}^2$.