Chapter 11, Problem 78GP

(a)

To determine

To Find: The formula for the ratio of wave speeds in the heavy section to that of the lighter section.

Answer to Problem 78GP

The formula for the ratio of wave speed in the heavy section to that in the lighter section is $\frac{v_H}{v_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}$ .

Explanation of Solution

Given:

There is a sin wave that is traveling down the stretched twoparts.

  $V_H$ and $V_L$ are the wave speeds in the heavy section and in the lighter section.

Formula used:

The speed of the wave s given as,

  $v=\sqrt{\frac{F_T}{\mu }}$

Where, $F_T$ is the tension force in the string and $\mu$ is represents the mass per unit length.

Calculation:

The tension inside the chord must be the same every place because the chord is not accelerating to the left or right. So, in both parts of the chords, the tension is the same. But there is a speed difference due to the different mass densities of both parts of the chord. Here, consider the $\mu$ as the mass per unit length of the chord’s each part. Then the ratio of speed is,

  $\frac{v_H}{v_L}=\frac{{\left(\sqrt{\frac{F_T}{\mu }}\right)}_H}{{\left(\sqrt{\frac{F_T}{\mu }}\right)}_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}$

Conclusion:

Hence, the formula for the ratio of wave speed in the heavy section to that in the lighter section is determined.

(b)

To determine

To Find: The ratio of the wavelengths in the two sections.

Answer to Problem 78GP

The ratio of the wavelengths in the two sections is $\sqrt{\frac{{\mu }_L}{{\mu }_H}}$ .

Explanation of Solution

Given:

There is a sine wave that is traveling down the stretched twoparts.

Here, $V_H$ and $V_L$ are the wave speeds in the heavy section and in the lighter section.

Formula used:

The ratio of the wave speeds in the two sections is given as,

  $\frac{v_H}{v_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}\mathrm{.....}\left(1\right)$

Where, ${\mu }_L$ and ${\mu }_H$ represent the mass per unit length in the heavy section and lighter section respectively.

And,

The wavelength is given as,

  $\lambda =\frac{v}{f}$

Where, v is the speed and $f$ is the frequency of the wave.

Calculation:

By using the wavelength formula in both sections, the ratio of the wavelength in the heavy and lighter section is given as,

  $\frac{{\lambda }_H}{{\lambda }_L}=\frac{{\left(\frac{v}{f}\right)}_H}{{\left(\frac{v}{f}\right)}_L}=\frac{v_H}{v_L}$

To remains continuous at the boundary, the chord must be having the same frequencies. By using the equation (1), the wavelength ratio becomes

  $\begin{array}{l}\frac{{\lambda }_H}{{\lambda }_L}=\frac{{\left(\frac{v}{f}\right)}_H}{{\left(\frac{v}{f}\right)}_L}=\frac{v_H}{v_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}\\ \Rightarrow \frac{{\lambda }_H}{{\lambda }_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}\end{array}$

Conclusion:

The wavelength ratio is $\sqrt{\frac{{\mu }_L}{{\mu }_H}}$ .

(c)

To determine

To Check: The wavelength is greater in which medium.

Answer to Problem 78GP

The wavelength is greater in the lighter portion of the chord.

Explanation of Solution

Given:

There is a sin wave that is traveling down the stretched twoparts.

Here, $V_H$ and $V_L$ are the wave speeds in the heavy section and in the lighter section.

Formula used:

The wavelength ratio in each section is given as,

  $\frac{{\lambda }_H}{{\lambda }_L}=\sqrt{\frac{{\mu }_L}{{\mu }_H}}$

Where, ${\mu }_L$ and ${\mu }_H$ represents the mass per unit length in the heavy section and lighter section respectively.

Calculation:

On the basis of the above formula, the wavelength is inversely varied with the $\mu$ . If ${\mu }_H$ is greater than the ${\mu }_L$ , then the wavelength in the heavy section is lower than that in the lighter section that is ${\lambda }_H<{\lambda }_L$ . Therefore, the wavelength in the lighter chord is greater.

Conclusion:

Hence, the wavelength is maximum in the lighter section of the chord.