A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass per unit length , , and respectively. The instrument is played by vibrating the strings by varying the free length in between the range and . It is found that in st

Question

A musical instrument is made using four different metal strings 1, 2, 3 and 4 with mass per unit length $μ$ , $2 μ$ , $3 μ$ and $4 μ$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_{0}$ and $2 L_{0}$ . It is found that in string-1 $(μ)$ 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. [2019]

	List-I		List-II
(I)	String-1 $(μ)$	(P)	1
(II)	String-2 $(2 μ)$	(Q)	$\frac{1}{2}$
(III)	String-3 $(3 μ)$	(R)	$\frac{1}{\sqrt{2}}$
(IV)	String-4 $(4 μ)$	(S)	$\frac{1}{\sqrt{3}}$
		(T)	$\frac{3}{16}$
		(U)	$\frac{1}{16}$

If the tension in each string is $T_{0}$ , the correct match for the highest fundamental frequency in $f_{0}$ units will be,

Smart Achievers · Accepted Answer

(3)

$Frequency, ν = \frac{1}{2 ℓ} \sqrt{\frac{T}{m}} for first mode of vibration$

For $2 L_{0}$ to be maximum, $' ℓ'$ should be minimum.

$String-1 f_{0} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{μ}}$

$String-2 f_{2} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{2 μ}} = \frac{f_{0}}{\sqrt{2}}$

$String-3 f_{3} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{4 μ}} = \frac{f_{0}}{\sqrt{3}}$

$String-4 f_{4} = \frac{1}{2 L_{0}} \sqrt{\frac{T_{0}}{4 μ}} = \frac{f_{0}}{2}$

Solution

1 I → Q, II → P, III → R, IV → T

2 I → Q, II → S, III → R, IV → P

3 I → P, II → R, III → S, IV → Q

4
I → P, II → Q, III → T, IV → S

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