Chapter 4.8, Problem 46E

To determine

To find: a model for simple harmonic motion satisfying the specified conditions.

Answer to Problem 46E

  $d=3\sin \left(\frac{\pi }{3}t\right)$

Explanation of Solution

Given information:

Displacement of the motion is 0, Amplitude is 3 meters and period is 6 seconds.

Calculation:

The equation for simple harmonic motion is of the form

  $d=a\sin \omega t\mathrm{............}(1)$

Where $a$ is amplitude and $\text{perriod}=\frac{2\pi }{\omega }$

Since the amplitude of the simple harmonic motion is 3 meters, therefore

  $a=3$

Also, since the period of the simple harmonic motion is 6 seconds, therefore

  $\begin{array}{c}\frac{2\pi }{\omega }=6\\ \omega =\frac{2\pi }{6}\\ =\frac{\pi }{3}\end{array}$

Put the values $a=3,\omega =\frac{\pi }{3}$ in equation (1)

  $d=3\sin \left(\frac{\pi }{3}t\right)$