Chapter 4.8, Problem 60E

a.

To determine

What is the amplitude?

Answer to Problem 60E

  $\boxed{A=4}$

Explanation of Solution

Given information:

The graph below shows the displacement of an object in simple harmonic motion.

  

What is the amplitude?

Calculation:

The difference between the maximum displacement and minimum displacement of the graph is twice the amplitude of function.

Amplitude is,

  $\begin{array}{l}a=\frac{1}{2}\left(Maxi,displacement-\min i,displacement\right)\\ =\frac{1}{2}\left[4-\left(-4\right)\right]\\ =\frac{1}{2}\left[8\right]\\ A=4\end{array}$

Hence the amplitude is $\boxed{A=4}$ .

b.

To determine

What is the period?

Answer to Problem 60E

  $\boxed{period=3}$

Explanation of Solution

Given information:

The graph below shows the displacement of an object in simple harmonic motion.

  

What is the period?

Calculation:

The given function completes one cycle between the times at which the two maximum displacement occur is $3$

Hence the period is $\boxed{period=3}$

c.

To determine

Is the equation of the simple harmonic motion of the form $d=a\sin \omega t,or,d=a\cos \omega t$ ?

Answer to Problem 60E

  $\boxed{d=a\cos \omega t}$

Explanation of Solution

Given information:

The graph below shows the displacement of an object in simple harmonic motion.

  

Is the equation of the simple harmonic motion of the form $d=a\sin \omega t,or,d=a\cos \omega t$ ?

Calculation:

As the maximum displacement is $d=4at0\sec ond$

  $\begin{array}{l}d\ne 0,for,t=0\\ however\\ d=0,for,t=0\\ d=a\sin \omega t\end{array}$

Hence the equation of the simple harmonic motion is of the form $\boxed{d=a\cos \omega t}$ .