A particle undergoing simple harmonic motion moves with an angular frequency of 4\text{ ra… — JAMB UTME

A particle undergoing simple harmonic motion moves with an angular frequency of $4\text{ rad/s}$ and an amplitude of $0.5\text{ m}$ . What is the maximum speed of the particle in $\text{m/s}$ ?

Answer: 2 m/s

Answer

The maximum speed of the particle is

2.0\text{ m/s}

The magnitude of velocity in simple harmonic motion varies with displacement

x

according to

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

. The speed reaches its maximum value when the particle passes through the equilibrium position (

x = 0

), giving

v_{\text{max}} = \omega A

. Substituting

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

gives

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

Step-by-Step Solution

Identify the given physical parameters.

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

These values define the speed profile of the simple harmonic oscillator.

Apply the SHM formula for maximum speed.

v_{\text{max}} = \omega A

Peak speed occurs at the equilibrium position where displacement is zero.

Substitute the values to calculate the maximum speed.

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

Multiplying angular frequency by amplitude yields the maximum linear velocity.

Key Concept

Maximum speed in Simple Harmonic Motion

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