91影视

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 $f_{nA}=\frac{n{V}_A}{位}\mathrm{where}位=\frac{2L}{n}$where

$f_{nA}=\frac{n}{2L}\sqrt{\frac{蟿}{渭}}$

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

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

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

$$\begin{gathered} 位=\frac{2L}{1} \\ =2L \\ {f}_{1A}=\frac{1}{2L}\sqrt{\frac{蟿}{渭}}.............(FirstharmonicfrequencyofA) \end{gathered}$$

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鈥檚 fourth harmonic frequency matches with A鈥檚 first harmonic frequency.

The second harmonic of string A is given at wavelength:

$$\begin{gathered} 位=L \\ {f}_{2,A}=\frac{1}{L}\sqrt{\frac{蟿}{渭}}................(secondresonantfrequencyofA) \end{gathered}$$

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

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

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

The third harmonic of string A is given at wavelength:

$$\begin{gathered} 位=\frac{3L}{2} \\ {f}_{3,A=}\frac{2}{3L}\sqrt{\frac{蟿}{渭}}.............(thirdresonantfrequencyofA) \end{gathered}$$

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 $f_{3,a}鈮�f_{n,B}$