Chapter 6.6, Problem 143E

To determine

To calculate:

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

  $d=16\cos \frac{\pi }{4}t$

Answer to Problem 143E

The maximum displacement and least possible value of $t$ at $d=0$ are $16$ and $2$ respectively.

Explanation of Solution

Given information:

  $d=16\cos \frac{\pi }{4}t$

Calculation:

Know that:

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

So, comparing the given equation $d=16\cos \frac{\pi }{4}t$ with standard simple harmonic motion equation $y=A\cos \omega t$ :

  $A=16$

Therefore, maximum displacement is $16$ .

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

  $d=0$

Put value of $d$ as:

  $16\cos \frac{\pi }{4}t=0$

Divide by $16$ :

  $\begin{array}{l}\frac{16}{16}\cos \frac{\pi }{4}t=0\\ \cos \frac{\pi }{4}t=0\end{array}$

Taking inverse cosine as:

  $\begin{array}{l}{\cos }^{-1}\cos \frac{\pi }{4}t={\cos }^{-1}0\\ \frac{\pi }{4}t=\frac{\pi }{2}\end{array}$

Multiple by $\frac{4}{\pi }$ :

  $\begin{array}{l}\frac{\pi }{4}\times \frac{4}{\pi }t=\frac{\pi }{2}\times \frac{4}{\pi }\\ t=2\end{array}$

Therefore, the least possible value of $t$ at $d=0$ is $t=2$ .