Chapter 18, Problem 16PQ

As in Figure P18.16, a simple harmonic oscillator is attached to a rope of linear mass density 5.4 × 10⁻² kg/m, creating a standing transverse wave. There is a 3.6-kg block hanging from the other end of the rope over a pulley. The oscillator has an angular frequency of 43.2 rad/s and an amplitude of 24.6 cm. a. What is the distance between adjacent nodes? b. If the angular frequency of the oscillator doubles, what happens to the distance between adjacent nodes? c. If the mass of the block is doubled instead, what happens to the distance between adjacent nodes? d. If the amplitude of the oscillator is doubled, what happens to the distance between adjacent nodes?

FIGURE P18.16

To determine

The distance between the adjacent nodes.

Answer to Problem 16PQ

The distance between the adjacent nodes is $\underset{\_}{1.9\text{m}}$.

Explanation of Solution

Given that the linear mass density of the rope is $5.4\times {10}^{-2}\text{kg}/\text{m}$, the mass of the block is $3.6\text{kg}$, frequency of the oscillator is $43.2\text{rad}/\text{s}$ and the amplitude of the harmonic waves is $24.6\text{cm}$.

Write the expression for the wavelength of the wave.

$\lambda =\frac{v}{f}$                                                                                                                         (I)

Here, $\lambda$ is the wavelength of the wave, $v$ is the speed of the wave, and $f$ is the frequency of the wave.

Write the expression for the speed of the wave.

$v=\sqrt{\frac{F_T}{\mu }}$ (II)

Here, $F_T$ is the force of tension on rope, $\mu$ is the linear mass density.

Write the equation to find the frequency of the wave.

$f=\frac{\omega }{2\pi }$ (III)

Here, $\omega$ is the angular frequency of the wave.

Use equation (III) and (II) in (I).

$\begin{array}{c}\lambda =\frac{1}{f}\sqrt{\frac{F_T}{\mu }}\\ =\frac{2\pi }{\omega }\sqrt{\frac{F_T}{\mu }}\end{array}$ (IV)

Write the expression for the tension force on the rope (it is equal to the weight of the block hanged).

  $F_T=mg$

Here, $g$ is the acceleration due to gravity.

Use above expression in equation (IV).

$\lambda =\frac{2\pi }{\omega }\sqrt{\frac{mg}{\mu }}$ (V)

The distance between the adjacent nodes is equal to half the wavelength of the wave.

  $d_{\text{nodes}}=\frac{1}{2}\lambda$

Rewrite above equation using equation (V).

$d_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{\omega }\sqrt{\frac{mg}{\mu }}\right)$                                                                                               (VI)

Conclusion:

Substitute $43.2\text{rad}/\text{s}$ for $\omega$, $3.6\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $5.4\times {10}^{-2}\text{kg}/\text{m}$ for $\mu$ in equation (VI) to get $d_{\text{nodes}}$.

  $\begin{array}{c}\lambda =\frac{1}{2}\left(\frac{2\pi }{43.2\text{rad}/\text{s}}\sqrt{\frac{\left(3.6\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{5.4\times {10}^{-2}\text{kg}/\text{m}}}\right)\\ =1.9\text{m}\end{array}$

Therefore, the distance between the adjacent nodes is $\underset{\_}{1.9\text{m}}$.

To determine

The distance between the adjacent nodes when the angular frequency is doubled.

Answer to Problem 16PQ

The distance between the adjacent nodes when the angular frequency is doubled is $\underset{\_}{0.93\text{m}}$.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  $d_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{\omega }\sqrt{\frac{mg}{\mu }}\right)$

From the above equation it is clear that wavelength is inversely proportional to angular frequency.

The angular frequency is doubled. Replace $\omega$ by $2\omega$ in equation (VI) to find the new distance ${d^{\prime }}_{\text{nodes}}$.

$\begin{array}{c}{d^{\prime }}_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{2\omega }\sqrt{\frac{mg}{\mu }}\right)\\ =\left(\frac{\pi }{2\omega }\sqrt{\frac{mg}{\mu }}\right)\end{array}$ (VII)

Conclusion:

Substitute $43.2\text{rad}/\text{s}$ for $\omega$, $3.6\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $5.4\times {10}^{-2}\text{kg}/\text{m}$ for $\mu$ in the equation (VII) to get ${d^{\prime }}_{\text{nodes}}$.

  $\begin{array}{c}{d^{\prime }}_{\text{nodes}}=\left(\frac{\pi }{2\left(43.2\text{rad}/\text{s}\right)}\sqrt{\frac{\left(3.6\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{5.4\times {10}^{-2}\text{kg}/\text{m}}}\right)\\ =0.93\text{m}\end{array}$

Therefore, distance between the adjacent nodes when the angular frequency is doubled is $\underset{\_}{0.93\text{m}}$.

To determine

The distance between the adjacent nodes if the mass of the block is doubled.

Answer to Problem 16PQ

The distance between the adjacent nodes if the mass of the block is doubled is $\underset{\_}{2.6\text{m}}$.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  $d_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{\omega }\sqrt{\frac{mg}{\mu }}\right)$

When the mass of the block is doubled, $m$ becomes $2m$.Replace $m$ by $2m$ in equation (VI) to find the new distance ${d^{″}}_{\text{nodes}}$.

$\begin{array}{c}{d^{″}}_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{\omega }\sqrt{\frac{2mg}{\mu }}\right)\\ =\left(\frac{4\pi }{\omega }\sqrt{\frac{mg}{\mu }}\right)\end{array}$ (VIII)

Conclusion:

Substitute $43.2\text{rad}/\text{s}$ for $\omega$, $3.6\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $5.4\times {10}^{-2}\text{kg}/\text{m}$ for $\mu$ in the equation (VIII) to get ${d^{″}}_{\text{nodes}}$.

  $\begin{array}{c}{d^{″}}_{\text{nodes}}=\left(\frac{4\pi }{43.2\text{rad}/\text{s}}\sqrt{\frac{\left(3.6\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{5.4\times {10}^{-4}\text{kg}/\text{m}}}\right)\\ =2.6\text{m}\end{array}$

Therefore, the distance between the adjacent nodes if the mass of the block is doubled is $\underset{\_}{2.6\text{m}}$.

To determine

The distance between the nodes if the amplitude of the oscillator is doubled.

Answer to Problem 16PQ

The distance between the nodes remains the same even if the amplitude of the oscillator is doubled.

Explanation of Solution

Equation (VI) gives the expression for the distance between the nodes.

  $d_{\text{nodes}}=\frac{1}{2}\left(\frac{2\pi }{\omega }\sqrt{\frac{mg}{\mu }}\right)$

The above equation is independent of the amplitude term. Thus, even if the amplitude of the oscillator is doubled, the distance between the nodes do not change.

Conclusion:

Therefore, the distance between the nodes remains the same even if the amplitude of the oscillator is doubled.