Chapter 15, Problem 3SP

(a)

To determine

The sketch of the fundamental standing wave for the pipe.

Answer to Problem 3SP

The sketch of the fundamental standing wave for the pipe is given in figure 1.

Explanation of Solution

Given info: The length of the pipe is $40\text{cm}$.

Following figure gives the standing wave pattern of the fundamental standing wave for the pipe whose length is $40\text{cm}$.

Figure 1

The nodes are the point where two waves cancel each other and no motion occurs. The anti-nodes are the points of maximum amplitude.

Conclusion:

Therefore, the sketch of the fundamental standing wave for the pipe is given in figure 1.

(b)

To determine

The wavelength of the sound wave that interfere to form the fundamental wave.

Answer to Problem 3SP

The wavelength of the sound wave that interfere to form the fundamental wave is $0.8\text{m}$.

Explanation of Solution

Write the expression to calculate the fundamental wavelength inside the pipe.

$\frac{\lambda }{2}=L$

Here,

$\lambda$ is the fundamental wavelength inside the pipe

L is the length of the pipe

Substitute $40\text{cm}$ for L in the above equation and rearrange to calculate $\lambda$.

$\begin{array}{c}\frac{\lambda }{2}=40\text{cm}\\ \lambda =2\left(40\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =0.8\text{m}\end{array}$

Conclusion:

Therefore, the wavelength of the sound wave that interfere to form the fundamental wave is $0.8\text{m}$.

(c)

To determine

The frequency of the sound wave.

Answer to Problem 3SP

The frequency of the sound wave is $425\text{Hz}$.

Explanation of Solution

Write the expression to calculate the speed of the sound wave.

$v=f\lambda$

Here,

$\lambda$ is the wavelength of the sound wave

f is the frequency of the sound wave

Substitute $340\text{m}/\text{s}$ for v and $0.8\text{m}$ for $\lambda$ in the above equation to calculate f.

$\begin{array}{c}f\left(0.8\text{m}\right)=340\text{m}/\text{s}\\ f=\frac{340\text{m}/\text{s}}{0.8\text{m}}\\ =425\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the sound wave is $425\text{Hz}$.

(d)

To determine

The change in frequency of the sound wave.

Answer to Problem 3SP

The change in frequency of the sound wave is $\text{+}12.5\text{Hz}$.

Explanation of Solution

Write the expression to calculate the speed of the sound wave.

$v^{\prime }=f^{\prime }\lambda$

Here,

$v^{\prime }$ is the new speed of the sound

$f^{\prime }$ is the new frequency of the sound wave

Substitute $350\text{m}/\text{s}$ for $v^{\prime }$ and $0.8\text{m}$ for $\lambda$ in the above equation to calculate $f^{\prime }$.

$\begin{array}{c}{f}^{\prime }\left(0.8\text{m}\right)=350\text{m}/\text{s}\\ f^{\prime }=\frac{350\text{m}/\text{s}}{0.8\text{m}}\\ =437.5\text{Hz}\end{array}$

Write the formula to calculate the frequency change for the sound wave.

$\Delta f=f^{\prime }-f$

Here,

$\Delta f$ is the frequency change

Substitute $437.5\text{Hz}$ for $f^{\prime }$ and $425\text{Hz}$ for f in the above equation to calculate $\Delta f$.

$\begin{array}{c}\Delta f=437.5\text{Hz}-425\text{Hz}\\ =\text{+}12.5\text{Hz}\end{array}$

Conclusion:

Therefore, the change in frequency of the sound wave is $\text{+}12.5\text{Hz}$.

(e)

To determine

The sketch of the second harmonic and its wavelength and frequency.

Answer to Problem 3SP

The sketch of the second harmonic is given in the figure 2 and its wavelength is $40\text{cm}$ and frequency is $850\text{Hz}$.

Explanation of Solution

For the second harmonic, the frequency is twice as that of the fundamental standing-wave and wavelength is half of that value of the fundamental standing wave. Therefore the frequency of the second harmonic wave is $850\text{Hz}$ and wavelength is $40\text{cm}$.

The sketch of the second harmonic wave is shown below.

Figure 2

Conclusion:

Therefore, the sketch of the second harmonic is given in the figure 2 and its wavelength is $40\text{cm}$ and frequency is $850\text{Hz}$.