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# Answer the following by appropriately matching the lists based on the information given in the paragraph.A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length μ, 2μ, 3μ and 4μ 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 List-II(I) String-1 (μ) (p) 1 (II) String-2 (2μ) (Q) 1/2 (III) String-3 (3μ) (R) $1/\sqrt 2$(IV) String-4 (4μ) (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 units will be :A. I →P, II →Q, III →T, IV →SB. I →P, II →R, III →S, IV →QC. I →Q, II →S, III →R, IV →PD. I →Q, II →P, III →R, IV →T

Last updated date: 13th Jun 2024
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Hint: When string is fixed from both ends nodes will be formed at both ends. When a string is only fixed from one end and set free at the other end node is formed at fixed end and antinode is formed at the free end. Since tension and linear mass density are given it is possible to find out the velocity of the standing wave in the string.

Formulas used:
$v = \sqrt {\dfrac{T}{\mu }}$
\eqalign{ & \dfrac{\lambda }{2} = {l_0} \cr & \lambda = 2{l_0} \cr}

When string is fixed at both ends initially one loop will be formed and hence wavelength of that loop be $\lambda$
For one loop distance is $\dfrac{\lambda }{2}$
So initially $\dfrac{\lambda }{2} = {l_0}$ , $\lambda = 2{l_0}$
Velocity of wave would be $\sqrt {\dfrac{T}{\mu }} {\text{ }}$
Where Tis tension and $\mu$ is linear mass density
Hence fundamental frequency for the first string will be
$v = \sqrt {\dfrac{T}{\mu }}$
\eqalign{ & \Rightarrow \lambda {f_0} = \sqrt {\dfrac{T}{\mu }} \cr & \Rightarrow {f_0} = \dfrac{1}{\lambda }\sqrt {\dfrac{T}{\mu }} \cr & \therefore {f_1} = {f_0} = \dfrac{1}{{2{L_0}}}\sqrt {\dfrac{T}{\mu }} \cr}
For the second string fundamental frequency will be
$v = \sqrt {\dfrac{T}{\mu }}$
\eqalign{ & \Rightarrow \lambda {f_2} = \sqrt {\dfrac{T}{{2\mu }}} \cr & \Rightarrow {f_2} = \dfrac{1}{\lambda }\sqrt {\dfrac{T}{\mu }} \cr & \Rightarrow {f_2} = \dfrac{1}{{2{L_0}}}\sqrt {\dfrac{T}{{2\mu }}} \cr & \therefore {f_2} = \dfrac{{{f_0}}}{{\sqrt 2 }} \cr}
For the third string fundamental frequency will be
$v = \sqrt {\dfrac{T}{\mu }}$
\eqalign{ & \Rightarrow \lambda {f_3} = \sqrt {\dfrac{T}{{3\mu }}} \cr & \Rightarrow {f_3} = \dfrac{1}{\lambda }\sqrt {\dfrac{T}{{3\mu }}} \cr & \Rightarrow {f_3} = \dfrac{1}{{2{L_0}}}\sqrt {\dfrac{T}{{3\mu }}} \cr & \therefore {f_3} = \dfrac{{{f_0}}}{{\sqrt 3 }} \cr}
For the fourth string fundamental frequency will be
$v = \sqrt {\dfrac{T}{\mu }}$
\eqalign{ & \Rightarrow \lambda {f_4} = \sqrt {\dfrac{T}{{4\mu }}} \cr & \Rightarrow {f_4} = \dfrac{1}{\lambda }\sqrt {\dfrac{T}{{4\mu }}} \cr & \Rightarrow {f_4} = \dfrac{1}{{2{L_0}}}\sqrt {\dfrac{T}{{4\mu }}} \cr & \therefore {f_4} = \dfrac{{{f_0}}}{2} \cr}
Hence
\eqalign{ & {f_1}:{f_2}:{f_3}:{f_4} = {f_0}:\dfrac{{{f_0}}}{{\sqrt 2 }}:\dfrac{{{f_0}}}{{\sqrt 3 }}:\dfrac{{{f_0}}}{2} \cr & \therefore {f_1}:{f_2}:{f_3}:{f_4} = 1:\dfrac{1}{{\sqrt 2 }}:\dfrac{1}{{\sqrt 3 }}:\dfrac{1}{2} \cr}
So the fundamental frequencies in the various strings will be in the given ratio. In all the strings the tension and length of vibrating ends is constant but mass density Is varying which results in variation of velocity.
Hence option B will be the answer.

Note: The velocities which we consider over here are velocities of waves in particular medium but not velocities of individual particles oscillating in that wave. Velocity of waves is constant as long as material density and source frequency remain constant but particle velocity at every position of wave is different. At maximum displacement position of wave, particle velocity there is minimum and at minimum displacement position particle velocity is maximum.
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