Question

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

Column 1	Column 2
String – 1	$1$
String – 2	$\frac{1}{2}$
String – 3	$\frac{1}{\sqrt{2}}$
String – 4	$\frac{1}{\sqrt{3}}$

Answer 1

Part A:

For string 1,

$T_0$ is the tension of the string

$L_0=$ Length of string

$\mu$ is the mass per unit length.
$f_0=$ Fundamental frequency

Maximum fundamental frequency when length minimum, $\mu =1$

$f_0=\frac{1}{L_0}\sqrt{\frac{T_0}{\mu }}$

Hence, A-I

Part B,

For string 2,

$\mu$is the mass per unit length.
$f_1=$Maximum fundamental frequency when length minimum, $\mu =2$

$f_1=\frac{1}{L_0}\sqrt{\frac{T_0}{2\mu }}$

$f_2=\frac{f_0}{\sqrt{2}}$

Hence, B-III

Part C,

For string 3,

$\mu$ is the mass per unit length.
$f_2=$ Maximum fundamental frequency when length minimum, $\mu =3$

$f_2=\frac{1}{L_0}\sqrt{\frac{T_0}{3\mu }}$

$f_2=\frac{f_0}{\sqrt{3}}$

Hence, C-IV

Part D,

For string 4,

$\mu$is the mass per unit length.
$f_2=$ Fundamental frequency

Maximum fundamental frequency when length minimum, $\mu =4$

$f_3=\frac{1}{L_0}\sqrt{\frac{T_0}{4\mu }}$

$f_3=\frac{f_0}{2}$

Hence, D-II
Hence, the correct answers are A-I, B-III, C-IV, and D-II.