Verified Solution • 2025–26 Session moderate

QJEEAdv19Ph55a1: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph55a1 Solution: A musical instrument is made using four different metal…

A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length μ\mu , 2 μ\mu , 3 μ\mu and 4 μ\mu respectively. The instrument is played by vibrating the strings by varying the free length in between the range L0 and 2L0. It is found that in string-1 μ\mu at free length L0 and tension T0 the fundamental mode frequency is f0.

List-I gives the above four strings while list-II lists the magnitude of some quantity.

JEE Advanced 2019 Paper 2 Offline Physics - Waves Question 25 English

The length of the strings 1, 2, 3 and 4 are kept fixed at L0, 3L02{{3{L_0}} \over 2} , 5L04{{5{L_0}} \over 4} and 7L04{{7{L_0}} \over 4} respectively. Strings 1, 2, 3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonies, respectively such that all the strings have same frequency.

The correct match for the tension in the four strings in the units of T0 will be

Multiple Choice Options

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

Step-by-Step Verified Solution

Case 1. L=L0, T=T0, f=f0L = {L_0},\,T = {T_0},\,f = {f_0}

f1=12L0T0μ{f_1} = {1 \over {2{L_0}}}\sqrt {{{{T_0}} \over \mu }}

Case 2. L=3L02L = {{3{L_0}} \over 2}

f2=32×3L02T22μ=f0{f_2} = {3 \over {2 \times {{3{L_0}} \over 2}}}\sqrt {{{{T_2}} \over {2\mu }}} = {f_0}

⇒f0=12L0T2μ⇒T2=T02\Rightarrow {f_0} = {1 \over {2{L_0}}}\sqrt {{{{T_2}} \over \mu }} \Rightarrow {T_2} = {{{T_0}} \over 2}

Case 3. L=5L04L = {{5{L_0}} \over 4}

f3=52×5L04T33μ=f0{f_3} = {5 \over {2 \times {{5{L_0}} \over 4}}}\sqrt {{{{T_3}} \over {3\mu }}} = {f_0}

⇒f0=23L0T3μ⇒T3=3T016\Rightarrow {f_0} = {2 \over {\sqrt 3 {L_0}}}\sqrt {{{{T_3}} \over \mu }} \Rightarrow {T_3} = {{3{T_0}} \over {16}}

Case 4. L=7L04L = {{7{L_0}} \over 4}

f4=142×7L04T44μ=f0{f_4} = {{14} \over {2 \times {{7{L_0}} \over 4}}}\sqrt {{{{T_4}} \over {4\mu }}} = {f_0}

⇒f0=2L0T4μ⇒T4=T016\Rightarrow {f_0} = {2 \over {{L_0}}}\sqrt {{{{T_4}} \over \mu }} \Rightarrow {T_4} = {{{T_0}} \over {16}}

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