Chapter 18, Problem 33P

(a)

To determine

The number of loops exhibit in the pattern.

Answer to Problem 33P

The number of loops exhibit in the pattern is $3\text{ loops}$.

Explanation of Solution

The expression for given wave function is,

  $y=0.00200\sin \left(\pi x\right)\cos \left(100\pi t\right)$

Here, $y$ is the position of the wave at y axis, $x$ is the position of the wave at x axis, and $t$ is the time period.

Compare the above equation with standard wave function.

  $y=2A\sin kx\cos \omega t$

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

Write the expression for wave number.

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

Here, $\lambda$ is the wavelength of the standing wave.

Write the expression for angular frequency.

  $\omega =2\pi f$                                                                                              (II)

Here, $f$ is the frequency of vibration.

Write the expression for distance between adjacent nodes.

  $d_{\text{NN}}=\frac{\lambda }{2}$                                                                                               (III)

Write the expression for number of loops formed on the string.

  $n=\frac{L}{d_{\text{NN}}}$                                                                                             (IV)

Here, $n$ is the number of loops and $L$ is the length of the thin wire.

Conclusion:

Substitute $\pi$ for $k$ in equation (I) to find $\lambda$.

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

Substitute $100\pi$ for $\omega$ in equation (II) to find $f$.

  $\begin{array}{c}100\pi =2\pi f\\ f=50.0\text{Hz}\end{array}$

Substitute $2.00\text{m}$ for $\lambda$ in equation (III) to find $d_{\text{NN}}$.

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

Substitute $3.00\text{m}$ for $L$ and $1.00\text{m}$ for $d_{\text{NN}}$ in equation (IV) to find $n$.

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

Therefore, the number of loops exhibit in the pattern is $3\text{ loops}$.

(b)

To determine

The fundamental frequency of vibration.

Answer to Problem 33P

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

Explanation of Solution

Write the expression for speed of the wave.

  $v=\frac{\omega }{k}$                                                                                                    (V)

Write the expression for wavelength of the standing wave vibration.

  $d_{\text{NN}}=\frac{{\lambda }_b}{2}$                                                                                               (VI)

Here, $d_{\text{NN}}$ is the distance between the adjacent nodes and ${\lambda }_b$ is the wavelength of the standing wave.

Write the expression for fundamental frequency.

  $f_b=\frac{v_a}{{\lambda }_b}$                                                                                                  (VII)

Here, $f_b$ is the fundamental frequency of the vibration and $v_a$ is the speed of the wave.

Conclusion:

Substitute $100\pi {\text{s}}^{-1}$ for $\omega$ and $\pi {\text{m}}^{-1}$ for $k$ in equation (VI) to find $v$.

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

Substitute $3.00\text{m}$ for $d_{\text{NN}}$ in equation (VI) to find ${\lambda }_b$.

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

Substitute $6.00\text{m}$ for ${\lambda }_b$ and $100\text{m/s}$ for $v_a$ in equation (VII) to find $f_b$.

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

Therefore, the fundamental frequency of vibration is $16.7\text{Hz}$.

(c)

To determine

The number of loops are present in the new pattern.

Answer to Problem 33P

The number of loops are present in the new pattern is one loop.

Explanation of Solution

Write the expression for tension in the string.

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

Here, $v_0$ is the increasing speed, $T_0$ is the tension in the string, and $\mu$ is the linear mass density of the wire.

Write the expression for tension increases in the string.

  $v_c=\sqrt{\frac{9T_0}{\mu }}$                                                                                      (IX)

Write the expression for constant wavelength.

  ${\lambda }_c=\frac{v_c}{f_a}$                                                                                            (X)

Here, ${\lambda }_c$ is the constant wavelength, $v_c$ is the increasing speed, and $f_a$ is the frequency.

Write the expression for wavelength of the standing wave vibration.

  $d_{\text{NN}}=\frac{{\lambda }_c}{2}$                                                                                        (XI)

Here, $d_{\text{NN}}$ is the distance between the adjacent nodes.

Conclusion:

Substitute $100\text{m/s}$ for $v_0$ in equation (IX) to find $v_c$.

  $\begin{array}{c}{v}_0=3\sqrt{\frac{T_0}{\mu }}\\ =3v_0\\ =3\left(100\text{m/s}\right)\\ =300\text{m/s}\end{array}$

Substitute $300\text{m/s}$ for $v_c$ and $50\text{Hz}$ for $f_a$ in equation (X) to find ${\lambda }_c$.

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

Substitute $6.00\text{m}$ for ${\lambda }_c$ in equation (XI) to find $d_{\text{NN}}$.

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

Hence, one loop formed in the string.