Chapter 13, Problem 52SGR

To determine

To find:Sine equation to represent the vibration of the trampoline $y$ as a function of time $t$ .

Answer to Problem 52SGR

  $y=5\sin 20\pi t$

Explanation of Solution

Given information:

Amplitude of the function $=a=5$ feet

Frequency $=10$ hertz

Concept:

If equation in the form

  $y=a\sin \left(bx\right)$

Then

Amplitude $=a$

Period $=\frac{2\pi }{\left|b\right|}$

Given

Amplitude of the function $=a=5$ feet

Frequency $=10$ hertz

We know

Period is the reciprocal of Frequency

Therefore,

Period $=\frac{1}{10}$

Formula for period is

Period $=\frac{2\pi }{\left|b\right|}$

Therefore,

  $\begin{array}{l}\Rightarrow \frac{2\pi }{\left|b\right|}=\frac{1}{10}\\ \Rightarrow \left|b\right|=2\pi \times 10\\ \Rightarrow \left|b\right|=20\pi \\ \Rightarrow b=\pm 20\pi \end{array}$

On substituting values in general equation

We get

  $\begin{array}{l}y=a\sin \left(bx\right)\\ \Rightarrow y=5\sin \left(\pm 20\pi t\right)\\ \Rightarrow y=5\sin 20\pi t\end{array}$

Therefore,

Required sine equation is

  $y=5\sin 20\pi t$