Question: Figure 17-30 shows a stretched string of lengthand pipes a,b, c, and dof lengthsL, 2L, L/2, andL/2, respectively. The string’s tension is adjusted until the speed of waves on the string equals the speed of sound waves in the air.Thefundamental mode of oscillationis then set up on the string. In which pipe will the sound produced by the string cause resonance, and what oscillation mode willthat sound set up?

1. The length of the pipe a is L
2. The length of the pipe b is 2L.
3. The length of the pipe c is$\frac{L}{2}$
4. The length of the pipe d is$\frac{L}{2}$
5. The speed of wave on the string equals to the speed of sound.
6. The fundamental mode of n = 1oscillation is set up on the string.

By using the formula for resonant frequencyf for the pipe in which one end is open and other is closed, and for the pipe in which both ends are open, find the corresponding frequencies in each pipe. Comparing these frequencies with the frequency of the string, find the pipe in which cause resonance by the sound produced by the string and the oscillation mode that the resonating sound will set up.

Formulae are as follow:

$f=\frac{nv}{4L}$

Where,

$n=1,3,5,\dots .$

$f=\frac{nv}{2L}$

Where,

$n=1,3,5,\dots .$

Where,

f is frequency, L is length and vis velocity.

The resonant frequencyforthe pipe in which one end is open and other is closed is given by,

$f=\frac{v}{\lambda }$

Where,v is the speed of sound and $\lambda$is the wavelength,

$\lambda =\frac{4L}{n}$

Where,L is the length of the pipe and$n=1,3,5,\dots .$

For fundamental mode n = 1

Therefore, the resonant frequency is given by,

$f=\frac{v}{4L}$

Also, the resonant frequency f for pipe in which both ends are open is given by,

$f=\frac{v}{\lambda }$

Where, v is the speed of sound and $\lambda$is the wavelength,

$\lambda =\frac{2L}{n}$

where, L is the length of the pipe and$n=1,2,3,\dots .$

For fundamental mode, n = 1

Therefore, the resonant frequency is given by,

$f=\frac{v}{2L}$

Let,$f_a,f_b,f_{c}and{f}_d$and be the frequencies in pipe a, b, c and d respectively.

Pipe ain whichone end is open and other is closed and L =L . So,

$f_a=\frac{v}{4L}$

Similarly, for pipe b with L 2 L ,

$f_a=\frac{v}{4\left(2L\right)}=\frac{v}{8L}$

For pipe c in which both ends are open and
$$\begin{gathered} L=\frac{L}{2}, \\ {f}_c=\frac{v}{2\left(\frac{L}{2}\right)} \\ {f}_c=\frac{v}{L} \end{gathered}$$

For pipe d in which one end is open andthe other is closed and
$$\begin{gathered} L=\frac{L}{2}, \\ {f}_d=\frac{v}{4\left(\frac{L}{2}\right)} \\ {f}_d=\frac{v}{2L} \end{gathered}$$

Now, the frequency of string with length for fundamental mode is given by,

$f_s=\frac{v}{2L}$

Where

$\lambda =2L$

Therefore,

$f_s=f_d$

Hence, the sound produced by the string will cause resonance in pipe D andthe fundamental mode of oscillation n = 1 will be set up by the sound.

Thus, by using the formula for resonant frequency for the pipe in which one end is open and other is closed and for the pipe in which both ends are open, this question can be solved.