The maximum acceleration of a body oscillating in simple harmonic motion is 8\text{ m/s}^2… — JAMB UTME

The maximum acceleration of a body oscillating in simple harmonic motion is $8\text{ m/s}^2$ . If the period of oscillation is $\pi\text{ s}$ , calculate the amplitude of the oscillation in meters.

Cevap: 2 m

Cevap

The amplitude of the oscillation is

2.0\text{ m}

The correct answer of

2.0\text{ m}

is obtained by first deriving the angular frequency

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

, and then using the relation

a_{\text{max}} = \omega^2 A

to solve for amplitude:

A = \frac{8}{2^2} = 2.0\text{ m}

Adım Adım Çözüm

Calculate angular frequency (

\omega

) from the given period (

T

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

Angular frequency specifies the rate of phase change in oscillations.

Apply the maximum acceleration formula for simple harmonic motion to determine amplitude (

A

a_{\text{max}} = \omega^2 A \implies 8 = 2^2 \times A \implies A = 2.0\text{ m}

In simple harmonic motion, maximum acceleration occurs at the extreme position and equals

\omega^2 A

Anahtar Kavram

Simple Harmonic Motion Acceleration and Period Relationship

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