Question

The free length of all four strings is varied from $L_0$ to 2$L_0$. Find the maximum fundamental frequency of 1, 2, 3, 4 in terms of $f_0$ (tension is same in all strings)

A. String(µ)-1	E. 1
B. String(2µ)-2	F. $\frac{1}{2}$
C. String(3µ)-3	G. $\frac{1}{\sqrt{2}}$
D. String(4µ)-4	H. $\frac{1}{\sqrt{3}}$
	I. $\frac{1}{16}$
	J. $\frac{3}{16}$

Answer 1

Hint: Remember the speed of transverse waves travelling in the string. On string transverse stationary waves are formed by the superposition of direct and the reflected transverse waves. Normal mode or fundamental frequency occurs when length is minimum. If you know all these points and the equation, we can easily find the solution.

Formula used:
Fundamental frequency can be calculated by the equation:
${{f}_{1}}={{f}_{0}}=\frac{1}{2{{L}_{0}}}\sqrt{\frac{{{T}_{0}}}{\mu }}$

Complete answer:
Waves travelling along a string superimpose and create standing waves. Fundamental frequency is the lowest possible frequency of wave in string. It happens at minimum length. The other natural frequencies with which standing waves form on string are classified into first overtone, second overtone, third overtone etc.

In general, any integral multiple of the fundamental frequency is allowed. An integral multiple of a frequency is called its harmonic. All overtones are harmonic of fundamental frequency and all harmonics of fundamental frequency are overtones.

Consider tension as T and length as $L_0$ and it is given that the tension is the same. Now, fundamental frequency can be calculated by the equation:
Fundamental frequency of string 1,
${{f}_{1}}={{f}_{0}}=\frac{1}{2{{L}_{0}}}\sqrt{\frac{{{T}_{0}}}{\mu }}$

Fundamental frequency of string 2,
${{f}_{2}}=\frac{1}{2{{L}_{0}}}\sqrt{\frac{{{T}_{0}}}{2\mu }}$

In terms of $f_0$, the fundamental frequency of string 2 in terms of $f_0$ is
${{f}_{2}}=\frac{{{f}_{0}}}{\sqrt{2}}$

Fundamental frequency of string 3,
${{f}_{3}}=\frac{1}{2{{L}_{0}}}\sqrt{\frac{{{T}_{0}}}{3\mu }}$

In terms of $f_0$, fundamental frequency of string 3 is
${{f}_{3}}=\frac{{{f}_{0}}}{\sqrt{3}}$

Fundamental frequency of string 4,
${{f}_{4}}=\frac{1}{2{{L}_{0}}}\sqrt{\frac{{{T}_{0}}}{4\mu }}$
In terms of $f_0$, fundamental frequency of string 4 is
${{f}_{4}}=\frac{1}{2}{{f}_{0}}$

Therefore, the answer is: $A\to E;B\to G;C\to H;D\to F$.

Note: In question it is given that the free length of all four strings is varied from $L_0$ to 2$L_0$ but we only take length as $L_0$ since we only wanted fundamental frequency. You should be careful taking length while doing this problem.