A body of mass 0.2\text{ kg} undergoes simple harmonic motion with an angular frequency of… — JAMB UTME

A body of mass $0.2\text{ kg}$ undergoes simple harmonic motion with an angular frequency of $10\text{ rad/s}$ and an amplitude of $0.04\text{ m}$ . What is the magnitude of the maximum restoring force acting on the body in newtons?

Answer: 0.8 N

Answer

The magnitude of the maximum restoring force acting on the body is

0.8\text{ N}

The maximum restoring force in simple harmonic motion occurs at maximum displacement (

y = A

) and is given by

F_{\text{max}} = m \omega^2 A

. Substituting

m = 0.2\text{ kg}

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

yields

F_{\text{max}} = 0.2 \times (10)^2 \times 0.04 = 0.8\text{ N}

Step-by-Step Solution

Identify the given physical quantities from the problem statement.

m = 0.2\text{ kg}

\omega = 10\text{ rad/s}

, and

A = 0.04\text{ m}

These parameters are required to calculate acceleration and restoring force in simple harmonic motion.

Calculate the maximum acceleration of the oscillating body.

a_{\text{max}} = \omega^2 A = (10)^2 \times 0.04 = 4.0\text{ m/s}^2

In simple harmonic motion, maximum acceleration occurs at maximum displacement (the amplitude).

Determine the maximum restoring force.

F_{\text{max}} = m a_{\text{max}} = 0.2 \times 4.0 = 0.8\text{ N}

According to Newton's second law, force is the product of mass and acceleration.

Key Concept

Maximum Restoring Force in Simple Harmonic Motion

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