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 $L_0$ and $2L_0$. It is found that in string- 1 $\left(\mu \right)$ 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)	$1/16$

The length of the strings $1,2,3$ and 4 are kept fixed at $L_0,\frac{3L_0}{2},\frac{5L_0}{4}$ and $\frac{7L_0}{4}$ respectively.
Strings $1,2,3$ and 4 are vibrated at their 1 st, 3 rd, 5 th, and 14 th harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of $T_0$ will be,

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

Answer 1

The correct option is A

$I\to P,II\to R,III\to T,IV\to U$

$\lambda =2L$
$f=\frac{V}{\lambda }=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
For string (1)
$f_0=\frac{1}{2L_0}\times \sqrt{\frac{T_1}{\mu }}=\frac{1}{2L}\sqrt{\frac{T_0}{\mu }}$

For string (2)
$f_0=\frac{3\times 2}{2\times 3L_0}\sqrt{\frac{T_2}{2\mu }}\Rightarrow T_2=\frac{T_0}{2}$

For string (3)
$f_0=\frac{5\times 2}{5L_0}\sqrt{\frac{T_3}{3\mu }}\Rightarrow T_3=\frac{3T_0}{16}$

For string (4)
$f_0=\frac{14\times 4}{2\times 7L_0}\sqrt{\frac{T_4}{4\mu }}\Rightarrow T_4=\frac{T_0}{16}$

$I\to P,II\to Q,III\to T,IV\to U$