A particle executes simple harmonic motion along a straight line with an angular frequency… — JAMB UTME

A particle executes simple harmonic motion along a straight line with an angular frequency of $6.0\text{ rad/s}$ . What is the magnitude of the acceleration of the particle, in $\text{m/s}^2$ , when its displacement from the mean position is $0.50\text{ m}$ ?

Answer: 18 m/s^2

Answer

18.0 m/s^2

The magnitude of acceleration in simple harmonic motion is calculated using

a = \omega^2 x

. Substituting

\omega = 6.0\text{ rad/s}

and

x = 0.50\text{ m}

gives

a = (6.0)^2 \times 0.50 = 36 \times 0.50 = 18.0\text{ m/s}^2

Step-by-Step Solution

Identify the formula relating acceleration to angular frequency and displacement in SHM.

a = \omega^2 x

In simple harmonic motion, the magnitude of acceleration is directly proportional to displacement from the equilibrium position.

Substitute the given physical values into the equation.

a = (6.0)^2 \times 0.50

The given values are angular frequency

\omega = 6.0\text{ rad/s}

and displacement

x = 0.50\text{ m}

Compute the numerical product.

a = 18.0\text{ m/s}^2

Squaring

6.0

yields

36

, and multiplying by

0.50

gives

18.0

Key Concept

Acceleration in Simple Harmonic Motion

Estimated Time:1m 0s

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