Q14.24(b):

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 (b) 3 cm

Given Variables & Initial Setup

The system describes Simple Harmonic Motion (SHM) defined by the following parameters:

• Amplitude ($A$) = $5\text{ cm} = 0.05\text{ m}$
• Period ($T$) = $0.2\text{ s}$
• Displacement ($x$) = $3\text{ cm} = 0.03\text{ m}$

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

$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\text{ rad/s} \approx 31.42\text{ rad/s}$

Step 1: Calculating Velocity ($v$)

The velocity of a particle in SHM as a function of displacement $x$ is given by [Derived from the conservation of energy in SHM]:

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

Substituting the given values:

$v = \pm (10\pi) \sqrt{(5)^2 - (3)^2}\text{ cm/s}$

$v = \pm 10\pi \sqrt{25 - 9} = \pm 10\pi \sqrt{16} = \pm 10\pi(4)$

$v = \pm 40\pi\text{ cm/s} \approx \pm 125.66\text{ cm/s}$

$v = \pm 1.26\text{ m/s}$

Step 2: Calculating Acceleration ($a$)

The acceleration of a particle in SHM is given by the differential equation of motion [Per Newton's Second Law and Hooke's Law]:

$a = -\omega^2 x$

Substituting the known values:

$a = -(10\pi)^2 \times (3\text{ cm})$

$a = -100\pi^2 \times 3$

$a = -300\pi^2\text{ cm/s}^2$

Using $\pi^2 \approx 9.87$:

$a \approx -300 \times 9.87 = -2961\text{ cm/s}^2 = -29.61\text{ m/s}^2$

Visual Representation of SHM Vectors

Summary Table

Final Solution: The velocity of the body is approximately $\pm 1.26\text{ m/s}$ and the acceleration is $-29.61\text{ m/s}^2$.