Chapter 8.5, Problem 11AYU

Rework Problem 9 under the same conditions, except that at time $t=0,$ the object is at its rest position and moving down.

$a=7;T=5\pi$ seconds

To determine

The function that relates the displacement $d$ of the object from its rest position after $t$ seconds and $a=7,T=5\pi$ seconds, where an object attached to a coiled spring is pulled down a distance $a$ from its rest position, and then released. Assuming that the motion is simple harmonic with period $T$ and the positive direction of the motion is up.

Answer to Problem 11AYU

Solution:

The function that relates the displacement $d$ of the object from its rest position after $t$ seconds is given by $d(t)=-7\sin \left(\frac{2}{5}t\right)$, where $a=7,T=5\pi$ seconds

Explanation of Solution

Given information:

An object attached to a coiled spring is pulled down a distance $a$ from its rest position, and then released. Assuming that the motion is simple harmonic with period $T$ and the positive direction of the motion is up.

$a=7,T=5\pi$ seconds

Explanation:

An object that moves on a coordinate axis so that the displacement $d$ from the rest position at time $t$ is given by either $d(t)=a\cos \left(\omega t\right)$ or $d(t)=a\sin \left(\omega t\right)$, where $a$ and $\omega >0$ are constants, moves with simple harmonic motion has amplitude $\left|a\right|$ and period $T=\frac{2\pi }{\omega }$

The motion of the object is simple harmonic.

An object attached to a coiled spring is pulled downand the positive direction of the motion is up, so distance $a=-7$ units from its rest position

The object is at rest position, when $t=0$, so it is easy to use sine function.

Here, $a=-7,T=5\pi$ seconds

To find $\omega$, use the formula $T=\frac{2\pi }{\omega }$

$\Rightarrow 5\pi =\frac{2\pi }{\omega }$

$\Rightarrow \omega =\frac{2}{5}$

As $\omega =\frac{2}{5}$ and $a=-7$, $d(t)=a\sin \left(\omega t\right)$ becomes $d(t)=-7\sin \left(\frac{2}{5}t\right)$

Therefore, the function that models the motion of the object is $d(t)=-7\sin \left(\frac{2}{5}t\right)$