A particle of mass 0.50\text{ kg} executes simple harmonic motion along a straight line. W… — JAMB UTME

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.06\text{ m}$ , its speed is $0.80\text{ m/s}$ and its potential energy is $0.09\text{ J}$ . What is the magnitude of the maximum acceleration of the particle in $\text{m/s}^2$ ?

Answer: 10 m/s^2

Answer

The magnitude of the maximum acceleration of the particle is

10\text{ m/s}^2

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

, the angular frequency

\omega

10\text{ rad/s}

. Using

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

, the amplitude

A

0.10\text{ m}

. Substituting these values into

a_{\max} = \omega^2 A

yields

10\text{ m/s}^2

Step-by-Step Solution

Find angular frequency

\omega

from potential energy.

\omega = 10\text{ rad/s}

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

0.09 = \frac{1}{2}(0.50)\omega^2 (0.06)^2 \implies 0.09 = 0.0009 \omega^2 \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}

Find amplitude

A

from speed.

A = 0.10\text{ m}

Using speed

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

0.80 = 10\sqrt{A^2 - 0.06^2} \implies 0.08 = \sqrt{A^2 - 0.0036} \implies 0.0064 = A^2 - 0.0036 \implies A^2 = 0.0100 \implies A = 0.10\text{ m}

Calculate maximum acceleration.

a_{\max} = 10\text{ m/s}^2

Using maximum acceleration formula

a_{\max} = \omega^2 A

a_{\max} = 100 \times 0.10 = 10\text{ m/s}^2

Key Concept

Simple Harmonic Motion Energy and Kinematic Relations

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