String is stretched between two clamps separated by distance L . String B, with the same linear density and under the same tension as string A, is stretched between two clamps separated by distance 4L. Consider the first eight harmonics of stringB. For which of these eight harmonics of B(if any) does the frequency match the frequency of (a) A’s first harmonic, (b) A’s second harmonic, and (c)A’s third harmonic?

Length of string A is L.

Length of string B is 4L.

We can find the frequencies of A at given harmonics and can match them with all eight harmonic frequencies of B by using the formula for frequency for n^th modes of vibration and can get the answers to the questions.

The $n^{th}$resonant frequency of string A is $f_{nA}=\frac{n{V}_A}{\lambda }\mathrm{where}\lambda =\frac{2L}{n}$where

$f_{nA}=\frac{n}{2L}\sqrt{\frac{\tau }{\mu }}$

String B has the resonant frequency$f_{nB}=\frac{n{V}_B}{\lambda }\mathrm{where}\lambda =2L_{B}ln,\mathrm{and}{L}_B=4L$ where

$$\begin{gathered} f_{nB}=\frac{\left(n{v}_B\right)}{2\left(4L\right)} \\ =\frac{n}{8L}\sqrt{\frac{\tau }{\mu }}........\left(1\right) \\ =\frac{1}{4}{f}_{(n,A)} \end{gathered}$$

Hence, the first harmonic of string A is given as:

$$\begin{gathered} \lambda =\frac{2L}{1} \\ =2L \\ {f}_{1A}=\frac{1}{2L}\sqrt{\frac{\tau }{\mu }}.............(FirstharmonicfrequencyofA) \end{gathered}$$

So, if we put n = 4 in frequency $f_{nB}$of B that is equation (1), we get the resonant frequency of B as:

$f_{1,A}=f_{4,B}$

So, we can say that B’s fourth harmonic frequency matches with A’s first harmonic frequency.

The second harmonic of string A is given at wavelength:

$$\begin{gathered} \lambda =L \\ {f}_{2,A}=\frac{1}{L}\sqrt{\frac{\tau }{\mu }}................(secondresonantfrequencyofA) \end{gathered}$$

If we put n = 8, in equation (1), we get the resonant frequency of B as:

$$\begin{gathered} f_{2,B}=\frac{1}{L}\sqrt{\frac{\tau }{\mu }} \\ i.e.f_{2,A}=f_{8,B} \end{gathered}$$

Therefore, the eighth harmonic of B’s matches with the A’s second harmonic.

The third harmonic of string A is given at wavelength:

$$\begin{gathered} \lambda =\frac{3L}{2} \\ {f}_{3,A=}\frac{2}{3L}\sqrt{\frac{\tau }{\mu }}.............(thirdresonantfrequencyofA) \end{gathered}$$

And n = 1, 2, 3, 4, 5, 6, 7, 8.

By putting all these eight values of n in$f_{n,B}$, it is observed that no harmonic frequency of B matches with the third harmonic of A.

Therefore, we can say that the third frequency of A does not match with any frequency of B $f_{3,a}\ne f_{n,B}$