Chapter 7, Problem 8P

(a)

To determine

To find: The points x at which the node is located.

Answer to Problem 8P

The node is located for $x=n\pi ,n\in \mathbb{N}$.

Explanation of Solution

Given:

A standing wave in a tube of length 37.7 ft is represented by the function $y\left(x,t\right)=0.3\cos \frac{1}{2}x\cos 50\pi t$.

Calculation:

Node occurs when $\cos \frac{1}{2}x=0$.

Solve the above equation for x as follows.

$\begin{array}{l}\cos \frac{1}{2}x=0\\ \frac{x}{2}=\frac{n\pi }{2},n\in \mathbb{N}\\ x=n\pi ,n\in \mathbb{N}\end{array}$

The length of the tube is 37.7 ft so the value of x must be less than 37.7 ft.

The end points are represented by $x=0\text{ and }x=37.7$.

At both the points the value of $\cos \frac{x}{2}\ne 0$.

Therefore, nodes are not located at any endpoints.

(b)

To determine

To find: The frequency at which the air vibrates for the points that are not nodes.

Answer to Problem 8P

Each point which is not a node vibrates with the frequency of 25 Hz.

Explanation of Solution

Compare the given equation $y\left(x,t\right)=0.3\cos \frac{1}{2}x\cos 50\pi t$ with the standard equation $y\left(x,t\right)=A\cos \alpha x\cos \beta t$ and obtain that $\beta =50\pi$.

Note that $\beta =\frac{2\pi }{\text{period}}$ and frequency is the reciprocal of the time period.

Therefore calculate the frequency as follows.

$\begin{array}{c}f=\frac{1}{\text{period}}\\ =\frac{1}{\frac{2\pi }{\beta }}\\ =\frac{50\pi }{2\pi }\\ =25\end{array}$

Therefore, each point which is not a node vibrates with the frequency of 25 Hz.