Chapter 14, Problem 60P

Review. For the arrangement shown in Figure P14.60, the inclined plane and the small pulley are frictionless; the string supports the object of mass M at the bottom of the plane; and the string has mass m. The system is in equilibrium, and the vertical part of the string has a length h. We wish to study standing waves set up in the vertical section of the string. (a) What analysis model describes the object of mass M? (b) What analysis model describes the waves on the vertical part of the string? (c) Find the tension in the string. (d) Model the shape of the string as one leg and the hypotenuse of a right triangle. Find the whole length of the string. (e) Find the mass per unit length of the string. (f) Find the speed of waves on the string. (g) Find the lowest frequency for a standing wave on the vertical section of the string. (h) Evaluate this result for M = 1.50 kg, m = 0.750 g, h = 0.500 m, and θ = 30.0°. (i) Find the numerical value for the lowest frequency for a standing wave on the sloped section of the string.

Figure P14.60

To determine

The analysis model

Answer to Problem 60P

The object is described using constant acceleration model.

Explanation of Solution

The mass of the object supported by the string is $M$, the mass of the entire string is $m$ and the standing waves are set up on the vertical portion of the string having a length $h$.

The object of mass $M$ experiences a motion due to the tension on the string and the force of gravity. If the net force acting on the object is not zero, then the object of mass $M$, then the net force acting on the object accelerates the object constantly in a straight line following the constant acceleration principle.

Conclusion:

Therefore, the object of mass $M$ can be described using constant acceleration model.

To determine

The analysis model

Answer to Problem 60P

The waves on the vertical part of the string can be described using the waves under boundary conditions model.

Explanation of Solution

The mass of the object supported by the string is $M$, the mass of the entire string is $m$ and the standing waves are set up on the vertical portion of the string having a length $h$.

The vertical portion of the string is fixed at both the ends hence the boundary conditions stand for the waves. Write the general equation for the wavelength on the string.

    ${\lambda }_n=\frac{2L}{n}$

Here, $L$ is the length of the string and $n=1,2,\mathrm{3...}$. The equation shows that the wavelength of the string varies with the value of $n$.

Conclusion:

Therefore, the waves on the vertical part of the string can be described using the waves under boundary conditions model.

To determine

The tension on the string

Answer to Problem 60P

The tension on the string is $Mg\sin \theta$.

Explanation of Solution

Figure.1 shows the arrangement of the mass-pulley system.

Write the equation for the net force acting on the mass.

    $F=T-Mg\sin \theta$        (I)

Here, $T$ is the tension on the string acting in direction as shown in the figure, $M$ is the mass of the block and $g$ is the acceleration due to gravity. Here, the force $Mg\sin \theta$ and the tension $T$ acting on the string is in direction opposite to each other that the difference between them is zero.

Conclusion:

Re-write the equation (I) such that the net force is zero.

    $\begin{array}{c}T-Mg\sin \theta =0\\ T=Mg\sin \theta \end{array}$        (II)

Therefore, the tension on the string is $Mg\sin \theta$.

To determine

The length of the string

Answer to Problem 60P

The length of the string is $h\left(\left(1+\sin \theta \right)/\sin \theta \right)$.

Explanation of Solution

Write the equation for the sine of the angle of inclination.

    $\sin \theta =\frac{h}{lengthofstringontheinclinedplane}$        (III)

Rewrite the equation (I) to find the equation for the $lengthofstringontheinclinedplane$.

    $lengthofstringontheinclinedplane=\frac{h}{\sin \theta }$        (IV)

Write the equation for the total length of the string.

    $L=h+\left(lengthofstringontheinclinedplane\right)$        (V)

Conclusion:

Substitute equation (II) in equation (III).

    $\begin{array}{c}L=h+\frac{h}{\sin \theta }\\ =\frac{h\sin \theta +h}{\sin \theta }\\ =h\left(\frac{1+\sin \theta }{\sin \theta }\right)\end{array}$        (VI)

Therefore, the length of the strings is $h\left(\left(1+\sin \theta \right)/\sin \theta \right)$.

To determine

The mass per unit length of the string

Answer to Problem 60P

The mass per unit length of the string is $\left(m\sin \theta /h\left(1+\sin \theta \right)\right)$.

Explanation of Solution

Write the equation for the mass per unit length of the string.

    $\mu =\frac{m}{L}$        (VII)

Here, $m$ is the mass of the string and $L$ is the length of the string.

Conclusion:

Substitute equation (V) in equation (VI).

    $\begin{array}{c}\mu =\frac{m}{h\left(\frac{1+\sin \theta }{\sin \theta }\right)}\\ =\frac{m\sin \theta }{h\left(1+\sin \theta \right)}\end{array}$        (VIII)

Therefore, the mass per unit length of the string is $\left(m\sin \theta /h\left(1+\sin \theta \right)\right)$

To determine

The speed of waves on the string

Answer to Problem 60P

The speed of waves on the string is $\sqrt{\left(Mgh/m\right)\left(1+\sin \theta \right)}$.

Explanation of Solution

Write the equation for the speed of the waves on the string.

    $v=\sqrt{\frac{T}{\mu }}$        (IX)

Here, $T$ is the tension on the string and $\mu$ is the mass per unit length of the string

Conclusion:

Substitute equation (II) and equation (VIII) in equation (IX).

    $\begin{array}{c}v=\sqrt{\frac{Mg\sin \theta }{\left[\frac{m\sin \theta }{h\left(1+\sin \theta \right)}\right]}}\\ =\sqrt{\frac{Mgh}{m}(1+\sin \theta )}\end{array}$        (X)

Therefore, the speed of the waves on the string is $\sqrt{\left(Mgh/m\right)\left(1+\sin \theta \right)}$.

To determine

The lowest frequency for a standing wave

Answer to Problem 60P

The lowest frequency for the standing waves is $\sqrt{\left(Mg/4mh\right)\left(1+\sin \theta \right)}$.

Explanation of Solution

The length $h$ of the vertical portion of the string vibrates as a single loop in the fundamental mode of vibration. Therefore, the distance between the two nodes is equal to the length $h$. The distance between the antinodes is also half the wavelength in the fundamental mode of vibration.

    $\begin{array}{c}{d}_{NN}=\frac{\lambda }{2}=h\\ \lambda =2h\end{array}$        (XI)

Write the equation for the frequency of the standing wave.

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

Here, $v$ is the speed of the sound waves and $\lambda$ is the wavelength of the sound.

Conclusion:

Substitute equation (X) and equation (XI) in equation (XII).

    $\begin{array}{c}f=\frac{1}{2h}\sqrt{\frac{Mgh}{m}(1+\sin \theta )}\\ =\sqrt{\frac{Mg}{4mh}(1+\sin \theta )}\end{array}$        (XIII)

Therefore, the lowest possible frequency of the standing waves is $\sqrt{\left(Mg/4mh\right)\left(1+\sin \theta \right)}$.

To determine

The lowest frequency for a set of given values

Answer to Problem 60P

The lowest frequency for the standing waves is $121\text{Hz}$.

Explanation of Solution

Write the equation for the lowest frequency of the standing wave from equation (XIII).

    $f=\sqrt{\frac{Mg}{4mh}(1+\sin \theta )}$

Conclusion:

Substitute $1.50\text{kg}$ for $M$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $0.750\times {10}^{-3}\text{kg}$ for $m$, $0.50\text{m}$ for $h$ and $30°$ for $\theta$.

    $\begin{array}{c}f=\sqrt{\frac{\left(1.50\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{4\left(0.750\times {10}^{-3}\text{kg}\right)\left(0.50\text{m}\right)}(1+\sin 30°)}\\ =\sqrt{\frac{22.05}{0.0015}}\\ =121\text{Hz}\end{array}$

Therefore, the lowest frequency of the standing waves is $121\text{Hz}$.

To determine

The lowest frequency on the sloped section

Answer to Problem 60P

The lowest frequency for the standing waves on the sloped section is $60.6\text{Hz}$.

Explanation of Solution

The wavelength of the fundamental mode of vibration is twice the length of the sloped section.

    $\lambda =\frac{2h}{\sin \theta }$        (XIV)

Here, $h$ is the length of the vertical portion of the string.

Substitute equation (X) and equation (XIV) in equation (XII).

    $\begin{array}{c}f=\frac{1}{\left(\frac{2h}{\sin \theta }\right)}\sqrt{\frac{Mgh}{m}(1+\sin \theta )}\\ =\sin \theta \sqrt{\frac{Mg}{4mh}(1+\sin \theta )}\end{array}$

Substitute $1.50\text{kg}$ for $M$, $9.80{\text{m/s}}^{\text{2}}$ for $g$, $0.750\times {10}^{-3}\text{kg}$ for $m$, $0.50\text{m}$ for $h$ and $30°$ for $\theta$.

    $\begin{array}{c}f=\sin 30°\sqrt{\frac{\left(1.50\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)}{4\left(0.750\times {10}^{-3}\text{kg}\right)\left(0.50\text{m}\right)}(1+\sin 30°)}\\ =\sin 30°\sqrt{\frac{22.05}{0.0015}}\\ =\sin 30°\left(121\text{Hz}\right)\\ =60.6\text{Hz}\end{array}$

Therefore, the lowest frequency of the standing waves is $60.6\text{Hz}$.