Chapter 7, Problem 7P

(a)

To determine

To find: The value of $\alpha$ from the figure.

Answer to Problem 7P

The values of $\alpha$ for the given figures are $1,2,3\text{ and 4}$ respectively.

Explanation of Solution

Given:

The string has length $\pi$.

The equation of the wave has the form $y=A\sin \alpha x$.

Calculation:

It is given that the equation of the wave has the form $y=A\sin \alpha x$.

For the first diagram it can be observed that the wave will repeat after one more such trough.

As the length of the string is $\pi$, the period of the wave in first diagram is $p_1=2\pi$.

The value of $\alpha$ is calculated as follows.

$\begin{array}{c}\alpha =\frac{2\pi }{p_1}\\ =\frac{2\pi }{2\pi }\\ =1\end{array}$

Similarly for the second diagram the period is $p_2=\pi$.

Calculate the value of $\alpha$ as follows.

$\begin{array}{c}\alpha =\frac{2\pi }{p_2}\\ =\frac{2\pi }{\pi }\\ =2\end{array}$

Similarly for the third diagram the period is $p_3=\frac{2\pi }{3}$.

Calculate the value of $\alpha$ as follows.

$\begin{array}{c}\alpha =\frac{2\pi }{p_3}\\ =\frac{2\pi }{\frac{2\pi }{3}}\\ =3\end{array}$

Similarly for the fourth diagram the period is $p_4=\frac{\pi }{2}$.

Calculate the value of $\alpha$ as follows.

$\begin{array}{c}\alpha =\frac{2\pi }{p_4}\\ =\frac{2\pi }{\frac{\pi }{2}}\\ =4\end{array}$

Hence the values of $\alpha$ are $1,2,3\text{ and 4}$ respectively.

(b)

To determine

To sketch: The rough graph of the standing waves for next two possible values of $\alpha$.

Explanation of Solution

From part (a) it can be observed that the values of $\alpha$ follow a pattern.

Specifically the values are in an arithmetic progression.

Therefore, it can be concluded that the next values of $\alpha$ will be 5 and 6.

The graph for $\alpha =5$ is as shown below in Figure 1.

From Figure 1 it can be noted that it has the vertical period $\frac{2\pi }{5}$.

The graph for $\alpha =6$ is as shown below in Figure 2.

From Figure 2 it can be noted that it has the vertical period $\frac{\pi }{3}$.

(c)

To determine

To find: The value of $\beta$ so that the equation $y=A\cos \beta t$ will represent the same wave as in part (a).

Answer to Problem 7P

The value of $\beta$ is $880\pi$.

Explanation of Solution

Given:

The value of time t is fixed and each point which is not a node vibrates with the frequency 440 Hz.

Calculation:

Note that the frequency is the reciprocal of the time period.

Therefore, the time period is $\frac{1}{440}$.

The value of $\beta$ is calculated as follows.

$\begin{array}{c}\beta =\frac{2\pi }{\text{time period}}\\ =\frac{2\pi }{\frac{1}{440}}\\ =880\pi \end{array}$

Therefore, the value required $\beta$ is $880\pi$.

(d)

To determine

To find: The equations in the form $y\left(x,t\right)=A\sin \alpha x\cos \beta t$ from the data obtained in the parts (a) to (c).

Answer to Problem 7P

The respective equations are as follows.

For $\alpha =1$ the equation is $y\left(x,t\right)=\sin x\cos 880\beta t$.

For $\alpha =2$ the equation is $y\left(x,t\right)=\sin 2x\cos 880\beta t$.

For $\alpha =3$ the equation is $y\left(x,t\right)=\sin 3x\cos 880\beta t$.

For $\alpha =4$ the equation is $y\left(x,t\right)=\sin 4x\cos 880\beta t$.

Explanation of Solution

Given:

The value of time t is fixed and each point which is not a node vibrates with the frequency 440 Hz.

Assume the value of A = 1.

Calculation:

From part (c) the value of $\beta =880\pi$.

Therefore, the equations will have the form $y\left(x,t\right)=\sin \alpha x\cos 880\beta t$.

Use the values of $\alpha$ from part (a) and obtain the different equations for the respective graphs as follows.

For $\alpha =1$ the equation is $y\left(x,t\right)=\sin x\cos 880\beta t$.

For $\alpha =2$ the equation is $y\left(x,t\right)=\sin 2x\cos 880\beta t$.

For $\alpha =3$ the equation is $y\left(x,t\right)=\sin 3x\cos 880\beta t$.

For $\alpha =4$ the equation is $y\left(x,t\right)=\sin 4x\cos 880\beta t$.