Chapter 14, Problem 28P

(a)

To determine

The number of loops in the pattern

Answer to Problem 28P

The pattern exhibits a total of $3$ loops.

Explanation of Solution

Write the expression for the wave function of a standing wave.

    $y=2A\sin kx\cos \omega t$        (I)

Here, $A$ is the amplitude of the standing wave, $k$ is the wave number and $\omega$ is the angular frequency of the standing wave.

Write the given equation for the wave function of standing waves

    $y=0.002\sin \left(\pi x\right)\cos \left(100\pi t\right)$        (II)

Compare equation (I) and equation (II) to find the value of wave number.

    $k=\pi {\text{m}}^{\text{-1}}$        (III)

Write the equation for the wave number.

    $k=\frac{2\pi }{\lambda }$        (IV)

Here, $\lambda$ is the wavelength of the standing wave. Compare equation (III) and equation (IV) to find the wavelength of the standing wave.

    $\begin{array}{c}\frac{2\pi }{\lambda }=\pi {\text{m}}^{\text{-1}}\\ \lambda =2.00\text{m}\end{array}$

Compare equation (I) and equation (II) to find the value of the angular frequency

    $\omega =100\pi {\text{s}}^{\text{-1}}$        (V)

Write the expression for the angular frequency.

    $\omega =2\pi f$        (VI)

Here, $f$ is the frequency of the standing wave. Compare the equation (V) and (VI) to find the frequency of the standing wave.

    $\begin{array}{c}2\pi f=100\pi {\text{s}}^{\text{-1}}\\ f=\frac{100\pi {\text{s}}^{\text{-1}}}{2\pi }\\ =50\text{Hz}\end{array}$

Conclusion:

The distance between two adjacent nodes is half the wavelength of the standing wave. Write the equation for the distance between two adjacent nodes.

    $d_{NN}=\frac{\lambda }{2}$

Substitute $2.00\text{m}$ for $\lambda$

    $\begin{array}{c}{d}_{NN}=\frac{2.00\text{m}}{2}\\ =1.00\text{m}\end{array}$

The number of loops is the ration between the length of the wire and the distance between the two adjacent nodes. Write the equation for the number of loops.

    $n=\frac{L}{d_{NN}}$

Here, $n$ is the number of loops, $L$ is the length of the wire and $d_{NN}$ is the distance between the two adjacent nodes. Substitute $3.00\text{m}$ for $L$ and $1.00\text{m}$ for $d_{NN}$.

    $\begin{array}{c}n=\frac{3.00\text{m}}{1.00\text{m}}\\ =3\text{loops}\end{array}$

Therefore, the pattern exhibits a total of $3$ loops.

(b)

To determine

The fundamental frequency of vibration

Answer to Problem 28P

The fundamental frequency of vibration of the wire is $16.7\text{Hz}$.

Explanation of Solution

Refer SUB-PART (a)

Conclusion:

Write the equation for the speed of the wave.

    $v=\frac{\omega }{k}$

Here, $\omega$ is the angular frequency of the standing wave and $k$ is the wave number.

Substitute $100\pi {\text{s}}^{\text{-1}}$ for $\omega$ and $\pi {\text{m}}^{\text{-1}}$ for $k$.

    $\begin{array}{c}v=\frac{100\pi {\text{s}}^{\text{-1}}}{\pi {\text{m}}^{\text{-1}}}\\ =100\text{m/s}\end{array}$        (VII)

The simplest standing wave has the distance of the wire as the distance between the adjacent nodes as they have only two nodes, one at each end.

    $d_{NN}=3.00\text{m}$        (VIII)

Also, the distance between the nodes at this case is half the wavelength of the simplest standing wave.

    $d_{NN}=\frac{{\lambda }_a}{2}$        (IX)

Here, ${\lambda }_a$ is the wavelength of the simplest standing wave. Compare equation (VII) and equation (VIII).

    $\begin{array}{c}\frac{{\lambda }_a}{2}=3.00\text{m}\\ =6.00\text{m}\end{array}$

Write the expression for the frequency of the simplest standing wave in this case.

    $f_a=\frac{v}{{\lambda }_a}$

Substitute $100\text{m/s}$ for $v$ and $6.00\text{m}$ for ${\lambda }_a$.

    $\begin{array}{c}{f}_a=\frac{100\text{m/s}}{6.00\text{m}}\\ =16.7\text{Hz}\end{array}$

(c)

To determine

The number of loops in the new pattern

Answer to Problem 28P

The new pattern has $1$ loop.

Explanation of Solution

Write the equation for the velocity in terms of the tension on the string.

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

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

Write the equation for the tension on the string when the tension is increased by $9$ times.

    $T_0=9T$        (XI)

Here, $T_0$ is the new tension.

Write the new equation for the speed of the wave when the tension is increased.

    $v_0=\sqrt{\frac{T_0}{\mu }}$        (XII)

Conclusion:

Substitute equation (XI) in equation (XII).

    $\begin{array}{c}{v}_0=\sqrt{\frac{9T}{\mu }}\\ =3\sqrt{\frac{T}{\mu }}\\ =3v\end{array}$

Substitute $100\text{m/s}$ for $v$ from equation (VII).

    $\begin{array}{c}{v}_0=3\left(100\text{m/s}\right)\\ =300\text{m/s}\end{array}$

Write the equation for the wavelength of the wave when the tension on the string is increased.

    ${\lambda }_0=\frac{v_0}{f}$

Here, $v_0$ is the speed of the wave after the tension is increased and $f$ is the frequency of the wave. Substitute $300\text{m/s}$ for $v_0$ and $50\text{Hz}$ for $f$.

    $\begin{array}{c}{\lambda }_0=\frac{300\text{m/s}}{50\text{Hz}}\\ =6.00\text{m}\end{array}$

The distance between the nodes at this case is half the wavelength of the standing wave.

    $d_{NN}=\frac{{\lambda }_0}{2}$

Substitute $6.00\text{m}$ for ${\lambda }_0$.

    $\begin{array}{c}{d}_{NN}=\frac{6.00\text{m}}{2}\\ =3.00\text{m}\end{array}$

Therefore, the distance between two adjacent nodes when the tension is increased is $3.00\text{m}$. This shows that the distance between two adjacent nodes is same as the length of the wire. Hence, there will be only one loop after the tension is increased by $9$ times.