Chapter 4, Problem 20CT

To determine

Find an equation for the simple harmonic motion.

Answer to Problem 20CT

  $\boxed{d=-6\cos \left(\pi t\right)}$

Explanation of Solution

Given information:

Write the equation for the simple harmonic motion of a ball on a spring that starts at its lowest point of $6$ inches below equilibrium, bounces to its maximum height of $6$ inches above equilibrium, and returns to its lowest point in a total of $2$ seconds.

Calculation:

As, we know that, a point that moves on a coordinate is said to be in simple harmonic motion. If its distance $d$ from the origin at time $t$ is given by either $d=a\sin \omega t$ or $d=a\cos \omega t$ , where $a$ and $\omega$ are real numbers such that $\omega \ne 0$ . The motion has amplitude $\left|a\right|$ , period $\frac{2\pi }{\omega }$ and frequency $\frac{\omega }{2\pi }$ .

As, the ball is at its lowest point, when $t=0$ we will use the equation:

  $d=-a\cos \omega$

As, the distance between the highest and the lowest point is $6$ in $a=6$ .

As, the ball returns to the highest point every $2$ seconds, period is $T=2s$ . So,

  $\begin{array}{l}T=\frac{2\pi }{\omega }\\ T=2\\ \omega =\pi rad/s\end{array}$

Now, we will substitute $\omega =\pi$ , $a=6$ in the above expression of simple harmonic motion:

  $\begin{array}{l}d=-a\cos \omega t\\ d=-6\cos \pi t\end{array}$

Hence, the equation that describes the motion of the bobber is $\boxed{d=-6\cos \left(\pi t\right)}$ .