Question

A musical instrument is made using four different metal strings. $1,2,3$ and 4 with mass per unit length $\mu ,2\mu ,3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0\text{and}2L_0\text{. It is found that in string-}1\text{(}\mu )$ at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0.$ List-I gives the above four strings while List-II lists the magnitude of some quantity.

	List-I		List-II
(i)	String-1$\left(\mu \right)$	(P)	$1$
(ii)	String-2$\left(2\mu \right)$	(Q)	$1/2$
(iii)	String- 3$\left(3\mu \right)$	(R)	$1/\sqrt{2}$
(iv)	String-4$\left(4\mu \right)$	(S)	$1/\sqrt{3}$
		(T)	$3/16$
		(U)	$1/16$

If the tension in each string is $T_0$, the correct match for the highest fundamental frequency in $f_0$ units will be,

• A. $I\to P,II\to R,III\to S,IV\to Q$
• B. $I\to Q,II\to P,III\to R,IV\to T$
• C. $I\to Q,II\to S,III\to R,IV\to P$
• D. $I\to P,II\to Q,III\to T,IV\to S$

Answer 1

The correct option is A

$I\to P,II\to R,III\to S,IV\to Q$

$\frac{\lambda }{2}=L\Rightarrow 2L$
$f=\frac{V}{\lambda }=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
(i) $f_0=\frac{1}{2L_0}\sqrt{\frac{T_0}{\mu }}$ $\Rightarrow f_o$

(ii) $f_1=\frac{1}{2L_0}\sqrt{\frac{T_0}{2\mu }}$ $\Rightarrow f_1=\frac{f_o}{\sqrt{2}}$

(iii) $f_2=\frac{1}{2L_0}\sqrt{\frac{T_u}{3\mu }}$ $\Rightarrow f_2=\frac{f_o}{\sqrt{3}}$

(iv) $f_3=\frac{1}{2L_0}\sqrt{\frac{T_0}{4\mu }}$ $\Rightarrow f_3=\frac{f_o}{2}$

For all highest fundamental is when length is $L_0$
Hence the correct option is $I\to P,II\to R,III\to S,IV\to Q$