Chapter 18, Problem 32PQ

(a)

To determine

The first two harmonic frequencies of vibration for the wire.

Answer to Problem 32PQ

The first two harmonic frequencies of vibration for the wire are $14.8\text{ Hz and }29.7\text{ Hz}$ .

Explanation of Solution

Write the expression for the natural frequencies of vibration of a wire fixed at both ends.

  $f_n=n\left(\frac{v}{2L}\right)n=1,2,3,\mathrm{...}$                                                                                       (I)

Here, $f_n$ is the frequency of $n^{\text{th}}$ harmonic, $L$ is the length of the wire and $v$ is the speed of wave.

Write the expression for the speed of wave in terms of tension in string and mass per unit length.

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

Here, $T$ is the tension in string and $\mu$ is the mass per unit length.

Write the equation for $\mu$ .

  $\mu =\frac{m}{L}$

Here, $m$ is mass of the string.

Put the above equation in equation (II).

  $v=\sqrt{\frac{T}{m/L}}$                                                                                                             (III)

Put equation (III) in (I) to get final expression for $f_n$ .

  $f_n=\frac{n}{2L}\sqrt{\frac{T}{m/L}}$                                                                                                       (IV)

Conclusion:

It is given that length of the wire is $1.50\text{ m}$, mass of the wire is $25.0\text{ g}$ and the tension in the wire is $33.0\text{ N}$ .

The first two harmonics are obtained when $n$ is taken values $1$ and $2$ .

Substitute $1$ for $n$, $1.50\text{ m}$ for $L$, $33.0\text{ N}$ for $T$ and $25.0\text{ g}$ for $m$ in equation (IV) to find the first harmonic.

  $\begin{array}{c}{f}_1=\frac{1}{2\left(1.50\text{ m}\right)}\sqrt{\frac{33.0\text{ N}}{25.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}/1.50\text{ m}}}\\ =14.8\text{ Hz}\end{array}$

Substitute $2$ for $n$, $1.50\text{ m}$ for $L$, $33.0\text{ N}$ for $T$ and $25.0\text{ g}$ for $m$ in equation (IV) to find the second harmonic.

  $\begin{array}{c}{f}_2=\frac{2}{2\left(1.50\text{ m}\right)}\sqrt{\frac{33.0\text{ N}}{25.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}/1.50\text{ m}}}\\ =29.7\text{ Hz}\end{array}$

Therefore, the first two harmonic frequencies of vibration for the wire are $14.8\text{ Hz and }29.7\text{ Hz}$ .

(b)

To determine

The possible harmonic modes of vibration of the wire.

Answer to Problem 32PQ

The possible harmonic modes of vibration of the wire are $n=5\text{ harmonic at }74\text{ Hz},n=10\text{ at }148\text{ Hz},n=15\text{ at }222\text{ Hz, etc}\text{.}$ .

Explanation of Solution

It is given that the wire has a mode at $30.0\text{ cm}$ from one end. This could be any mode that has the node at $30.0\text{ cm}$ . Take $D=30.0\text{ cm}$ to be the distance between the adjacent nodes.

Write the expression for $D$ .

  $D=\frac{\lambda }{2}$

Here, $\lambda$ is the wavelength and $D$ is the distance between the adjacent nodes.

Rewrite the above equation for $\lambda$ .

  $\lambda =2D$                                                                                                                    (V)

Write the expression for the wavelength of harmonics.

  $\lambda =2\frac{L}{n}$

Rewrite the above equation for $n$ .

  $n=\frac{2L}{\lambda }$                                                                                                                  (VI)

Put equation (V) in equation (VI).

  $\begin{array}{c}n=\frac{2L}{2D}\\ =\frac{L}{D}\end{array}$

Substitute $1.50\text{ m}$ for $L$ and $30.0\text{ cm}$ for $D$ in the above equation to find the value of $n$ .

  $\begin{array}{c}n=\frac{1.50\text{ m}}{30.0\text{ cm}\frac{1\text{ m}}{100\text{ cm}}}\\ =\frac{1.50\text{ m}}{0.300\text{ m}}\\ =5\end{array}$

Thus the mode corresponds to the $5^{\text{th}}$ harmonic.

Substitute $5$ for $n$, $1.50\text{ m}$ for $L$, $33.0\text{ N}$ for $T$ and $25.0\text{ g}$ for $m$ in equation (IV) to find the frequency of the $5^{\text{th}}$ harmonic.

  $\begin{array}{c}{f}_5=\frac{5}{2\left(1.50\text{ m}\right)}\sqrt{\frac{33.0\text{ N}}{25.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}/1.50\text{ m}}}\\ =74\text{ Hz}\end{array}$

There could be also an entire wavelength fitting between the nodes at the end at $30.0\text{ cm}$ in which case $D=\lambda$ . Any half number of wavelengths could fit between these two points and still have nodes at both.

Write the expression for $D$ .

  $D=N\frac{\lambda }{2}$

Here, $N$ is any integer number of half wavelengths between the two points on the string.

Rearrange the above equation for $\lambda$ .

  $\lambda =\frac{2D}{N}$

Put the above equation in equation (VI).

  $\begin{array}{c}n=\frac{2L}{2D/N}\\ =\frac{2NL}{2D}\\ =N\frac{L}{D}\end{array}$

Substitute $1.50\text{ m}$ for $L$ and $30.0\text{ cm}$ for $D$ in the above equation to find the expression for $n$ .

  $\begin{array}{c}n=N\frac{1.50\text{ m}}{30.0\text{ cm}\frac{1\text{ m}}{100\text{ cm}}}\\ =N\frac{1.50\text{ m}}{0.300\text{ m}}\\ =5N\end{array}$

The expression for $n$ implies that the harmonics that satisfy this condition are the $5^{\text{th}},{10}^{\text{th}},{15}^{\text{th}}$ and so on.

Conclusion:

Substitute $10$ for $n$, $1.50\text{ m}$ for $L$, $33.0\text{ N}$ for $T$ and $25.0\text{ g}$ for $m$ in equation (IV) to find the frequency of the ${10}^{\text{th}}$ harmonic.

  $\begin{array}{c}{f}_{10}=\frac{10}{2\left(1.50\text{ m}\right)}\sqrt{\frac{33.0\text{ N}}{25.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}/1.50\text{ m}}}\\ =148\text{ Hz}\end{array}$

Substitute $15$ for $n$, $1.50\text{ m}$ for $L$, $33.0\text{ N}$ for $T$ and $25.0\text{ g}$ for $m$ in equation (IV) to find the frequency of the ${15}^{\text{th}}$ harmonic.

  $\begin{array}{c}{f}_{15}=\frac{15}{2\left(1.50\text{ m}\right)}\sqrt{\frac{33.0\text{ N}}{25.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}/1.50\text{ m}}}\\ =222\text{ Hz}\end{array}$

Therefore, the possible harmonic modes of vibration of the wire are $n=5\text{ harmonic at }74\text{ Hz},n=10\text{ at }148\text{ Hz},n=15\text{ at }222\text{ Hz, etc}\text{.}$ .