Q14.24(a):

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 (a) 5 cm

Step 1: Identifying Given Parameters and Physical Constants

Based on the provided motion parameters, we identify the following variables:

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

Step 2: Calculating Angular Frequency ($\omega$)

The angular frequency is related to the period of oscillation by the relation $\omega = \frac{2\pi}{T}$ [Per the definition of SHM periodicity].

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

Step 3: Calculating Velocity ($v$)

The velocity of a particle in Simple Harmonic Motion is given by the expression $v = \omega \sqrt{A^2 - x^2}$ [By the conservation of energy in SHM].

Substituting the given values:

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

$v = 10\pi \sqrt{0.0025 - 0.0025} = 0 \text{ m/s}$

Justification: At the extreme position ($x = A$), the body momentarily comes to rest before reversing its direction.

Step 4: Calculating Acceleration ($a$)

The acceleration of a particle in SHM is defined by $a = -\omega^2 x$ [Per the restoring force law $F = -kx$ and Newton's Second Law $F=ma$].

Substituting the values:

$a = -(10\pi)^2 \times (0.05)$

$a = -(100\pi^2) \times (0.05)$

$a = -5\pi^2 \text{ m/s}^2$

Using $\pi^2 \approx 9.87$:

$a \approx -5 \times 9.87 = -49.35 \text{ m/s}^2$

Visual Representation of SHM Extreme Position

Final Solution: At a displacement of 5 cm, the velocity is 0 m/s and the acceleration is $-5\pi^2 \text{ m/s}^2 \approx -49.35 \text{ m/s}^2$.