Chapter 11, Problem 90P

To determine

The amplitude satisfy the equation $A^{\prime }=2A\cos \left(\omega t\right)$.

Answer to Problem 90P

It is proved that amplitude satisfy the equation $A^{\prime }=2A\cos \left(\omega t\right)$, where, $A^{\prime }$ is the amplitude of the wave plotted.

Explanation of Solution

Consider the expression for $y$ as given below

$\begin{array}{c}y=A\left[\sin \left(kx-\omega t\right)+\sin \left(kx+\omega t\right)\right]\\ =A\sin \left(kx-\omega t\right)+A\sin \left(kx+\omega t\right)\end{array}$ (I)

Use the trigonometric relation given below to simplify equation (I)

$\sin \alpha +\sin \beta =2\sin \left[\frac{\left(\alpha +\beta \right)}{2}\right]\cos \left[\frac{\left(\alpha -\beta \right)}{2}\right]$ (II)

Use equation (II) in (I)

$\begin{array}{c}y=A\sin \left(kx-\omega t\right)+A\sin \left(kx+\omega t\right)\\ =2A\sin \left(\frac{kx-\omega t+kx+\omega t}{2}\right)\cos \left(\frac{kx-\omega t-(kx+\omega t)}{2}\right)\\ =2A\sin \left(kx\right)\cos \left(-\omega t\right)\\ =\left[2A\cos \left(\omega t\right)\right]\sin \left(kx\right)\end{array}$ (III)

Substitute $A^{\prime }$ for $2A\cos \left(\omega t\right)$ in equation (III)

$y=A^{\prime }\sin \left(kx\right)$ (IV)

Conclusion:

Let us check whether $2A\cos \left(-\omega t\right)$ holds each values of amplitudes of the graph

Substitute $5.0\text{cm}$ for $A$, $\frac{\pi }{5.0}\text{rad/cm}$ for $k$, $\frac{\pi }{6.0}\text{rad/s}$ in equation (II)

$\begin{array}{c}y\left(x,t\right)=\left(5.0\text{cm}\right)\left\{\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)x-\left(\frac{\pi }{6.0}\text{rad/s}\right)t\right)+\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)x+\left(\frac{\pi }{6.0}\text{rad/s}\right)t\right)\right\}\\ =2\left(5.0\text{cm}\right)\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)x\right)\cos \left(-\left(\frac{\pi }{6.0}\text{rad/s}\right)t\right)\\ =\left(10.0\text{cm}\right)\cos \left[\left(\frac{\pi }{6.0}\text{rad/s}\right)t\right]\sin \left[\left(\frac{\pi }{5.0}\text{rad/cm}\right)x\right]\end{array}$

Thus, the maximum value of $y$ is $10.0\text{cm}$ for $x=2.5\text{cm}$, and $t=0$.

For $t=1.0\text{s}$, the value of $y\left(x,t\right)$ is

$\begin{array}{c}y\left(2.5\text{cm},1.0\text{s}\right)=\left(10.0\text{cm}\right)\cos \left[\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(1.0\text{s}\right)\right]\sin \left[\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)\right]\\ =\left(10.0\text{cm}\right)\cos \left[\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(1.0\text{s}\right)\right]\times 1\\ =8.7\text{cm}\end{array}$

For $t=2.0\text{s}$

$\begin{array}{c}y\left(2.5\text{cm},1.0\text{s}\right)=\left(10.0\text{cm}\right)\cos \left[\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(2.0\text{s}\right)\right]\sin \left[\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)\right]\\ =\left(10.0\text{cm}\right)\cos \left[\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(2.0\text{s}\right)\right]\times 1\\ =5.0\text{cm}\end{array}$

These values are correctly matching with those in graph.

In case of original function, for $t=1.0\text{s}$

$\begin{array}{c}y=\left(5.0\text{cm}\right)\left[\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)-\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(1.0\text{s}\right)\right)+\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)+\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(1.0\text{s}\right)\right)\right]\\ =8.7\text{cm}\end{array}$

For $t=2.0\text{s}$

$\begin{array}{c}y=\left(5.0\text{cm}\right)\left[\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)-\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(2.0\text{s}\right)\right)+\sin \left(\left(\frac{\pi }{5.0}\text{rad/cm}\right)\left(2.5\text{cm}\right)+\left(\frac{\pi }{6.0}\text{rad/s}\right)\left(2.0\text{s}\right)\right)\right]\\ =5.0\text{cm}\end{array}$

Therefore amplitudes in graph satisfies equation $A^{\prime }=2A\cos \left(\omega t\right)$, where $A^{\prime }$ is the amplitude of the wave plotted