A particle executing simple harmonic motion has a maximum speed of 3.0\text{ m/s} and a ma… — JAMB UTME

A particle executing simple harmonic motion has a maximum speed of $3.0\text{ m/s}$ and a maximum acceleration of $12.0\text{ m/s}^2$ . What is the period of oscillation of the particle?

A
$\frac{\pi}{4}\text{ s}$
$\frac{\pi}{2}\text{ s}$ Answer
C
$\pi\text{ s}$
D
$2\pi\text{ s}$

Answer

The period of oscillation of the particle is

\frac{\pi}{2}\text{ s}

For simple harmonic motion, maximum speed is

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

and maximum acceleration is

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

. Dividing the maximum acceleration by the maximum speed gives

\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{12.0}{3.0} = 4.0\text{ rad/s}

. Using the formula for the period

T = \frac{2\pi}{\omega}

, we find

T = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

Step-by-Step Solution

Relate maximum speed and maximum acceleration to angular frequency

\omega = \frac{a_{\text{max}}}{v_{\text{max}}}

Since

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

and

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

, dividing

a_{\text{max}}

v_{\text{max}}

eliminates the amplitude

A

and gives

\omega

Calculate the angular frequency

\omega

\omega = \frac{12.0\text{ m/s}^2}{3.0\text{ m/s}} = 4.0\text{ rad/s}

Substitute the given numerical values into the expression for angular frequency.

Calculate the period

T

T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

The period of simple harmonic motion is related to angular frequency by

T = \frac{2\pi}{\omega}

Key Concept

Relationship between maximum velocity, maximum acceleration, angular frequency, and period in simple harmonic motion.

Estimated Time:1m 15s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic