Chapter 6.6, Problem 144E

To determine

To calculate:

The maximum displacement and least possible value of $t$ at $d=0$ for the simple harmonic motion equation:

  $d=\frac{1}{12}\sin 60\pi t$

Answer to Problem 144E

The maximum displacement and least possible value of $t$ at $d=0$ are $\frac{1}{12}$ and $\frac{1}{60}$ respectively.

Explanation of Solution

Given information:

  $d=\frac{1}{12}\sin 60\pi t$

Calculation:

Know that:

For the simple harmonic motion equation $y=A\sin \omega t$ , the maximum displacement is represented by $A$ .

So, comparing the given equation $d=\frac{1}{12}\sin 60\pi t$ with standard simple harmonic motion equation $y=A\sin \omega t$ :

  $A=\frac{1}{12}$

Therefore, maximum displacement is $\frac{1}{12}$ .

Now, the least possible value of $t$ at $d=0$ :

  $d=0$

Put value of $d$ as:

  $\frac{1}{12}\sin 60\pi t=0$

Multiple by $12$ :

  $\begin{array}{l}\frac{1}{12}\times 12\sin 60\pi t=12\times 0\\ \sin 60\pi t=0\end{array}$

Taking inverse sine as:

  $\begin{array}{l}{\sin }^{-1}\sin 60\pi t={\sin }^{-1}0\\ 60\pi t=\pi \end{array}$

Divide by $60\pi$ :

  $\begin{array}{l}\frac{60\pi }{60\pi }t=\frac{\pi }{60\pi }\\ t=\frac{1}{60}\end{array}$

Therefore, the least possible value of $t$ at $d=0$ is $t=\frac{1}{60}$ .