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PhyHardAdv 2019 • Paper 2

Tension Matching for Vibrating Strings at Equal Harmonic Frequency

Comprehension Passage

A musical instrument is constructed using four distinct metal strings, numbered 1, 2, 3, and 4, with mass per unit length given as μ\mu, 2μ2\mu, 3μ3\mu, and 4μ4\mu respectively. The instrument is played by vibrating the strings while adjusting the free length within the range between L0L_0 and 2L02L_0. It is observed that for string-1 (μ\mu) at a free length L0L_0 and tension T0T_0, the fundamental mode frequency is f0f_0.

List-I lists the four strings, and List-II contains the magnitude of a particular quantity.

List-IList-II(I) String-1 (μ)(P) 1(II) String-2 (2μ)(Q) 1/2(III) String-3 (3μ)(R) 1/2(IV) String-4 (4μ)(S) 1/3(T) 3/16(U) 1/16\begin{array}{ll} \text{List-I} & \text{List-II} \\ \text{(I) String-1 } (\mu) & \text{(P) } 1 \\ \text{(II) String-2 } (2\mu) & \text{(Q) } 1/2 \\ \text{(III) String-3 } (3\mu) & \text{(R) } 1/\sqrt{2} \\ \text{(IV) String-4 } (4\mu) & \text{(S) } 1/\sqrt{3} \\ & \text{(T) } 3/16 \\ & \text{(U) } 1/16 \end{array}

The lengths of strings 1, 2, 3, and 4 are fixed at L0L_0, 3L02\frac{3L_0}{2}, 5L04\frac{5L_0}{4}, and 7L04\frac{7L_0}{4} respectively. The strings 1, 2, 3, and 4 are made to vibrate at their 1st1^{\text{st}}, 3rd3^{\text{rd}}, 5th5^{\text{th}}, and 14th14^{\text{th}} harmonics respectively, such that all four strings vibrate at the same frequency. Choose the correct match for the tension in each of the four strings expressed in units of T0T_0.

Options

A

I →\rightarrow P, II →\rightarrow R, III →\rightarrow T, IV →\rightarrow U

B

I →\rightarrow P, II →\rightarrow Q, III →\rightarrow T, IV →\rightarrow U

Correct
C

I →\rightarrow P, II →\rightarrow Q, III →\rightarrow R, IV →\rightarrow T

D

I →\rightarrow T, II →\rightarrow Q, III →\rightarrow R, IV →\rightarrow U

Topics & Concepts

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Step-by-Step Solution

The fundamental frequency f0f_0 of String-1 at length L0L_0, mass per unit length μ\mu, and tension T0T_0 is given by: f0=12L0T0μf_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

In general, the frequency ff of the nthn^{\text{th}} harmonic for a string of length LL, mass per unit length μ′\mu', and tension TT fixed at both ends is: f=n2LTμ′f = \frac{n}{2L} \sqrt{\frac{T}{\mu'}}

We are given that all four strings vibrate at the same frequency f=f0f = f_0.

1. String-1 (I)

  • Mass per unit length: μ1=μ\mu_1 = \mu
  • Length: L1=L0L_1 = L_0
  • Harmonic: n1=1n_1 = 1

f1=12L0T1μ=f0=12L0T0μf_1 = \frac{1}{2L_0} \sqrt{\frac{T_1}{\mu}} = f_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

Equating the two expressions gives: T1=T0  ⟹  T1T0=1(P)T_1 = T_0 \implies \frac{T_1}{T_0} = 1 \quad \text{(P)}

2. String-2 (II)

  • Mass per unit length: μ2=2μ\mu_2 = 2\mu
  • Length: L2=3L02L_2 = \frac{3L_0}{2}
  • Harmonic: n2=3n_2 = 3

f2=32(3L02)T22μ=1L0T22μf_2 = \frac{3}{2 \left(\frac{3L_0}{2}\right)} \sqrt{\frac{T_2}{2\mu}} = \frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}}

Since f2=f0f_2 = f_0: 1L0T22μ=12L0T0μ\frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

T22μ=12T0μ\sqrt{\frac{T_2}{2\mu}} = \frac{1}{2} \sqrt{\frac{T_0}{\mu}}

Squaring both sides: T22μ=14T0μ  ⟹  T2=12T0  ⟹  T2T0=12(Q)\frac{T_2}{2\mu} = \frac{1}{4} \frac{T_0}{\mu} \implies T_2 = \frac{1}{2} T_0 \implies \frac{T_2}{T_0} = \frac{1}{2} \quad \text{(Q)}

3. String-3 (III)

  • Mass per unit length: μ3=3μ\mu_3 = 3\mu
  • Length: L3=5L04L_3 = \frac{5L_0}{4}
  • Harmonic: n3=5n_3 = 5

f3=52(5L04)T33μ=2L0T33μf_3 = \frac{5}{2 \left(\frac{5L_0}{4}\right)} \sqrt{\frac{T_3}{3\mu}} = \frac{2}{L_0} \sqrt{\frac{T_3}{3\mu}}

Since f3=f0f_3 = f_0: 2L0T33μ=12L0T0μ\frac{2}{L_0} \sqrt{\frac{T_3}{3\mu}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

T33μ=14T0μ\sqrt{\frac{T_3}{3\mu}} = \frac{1}{4} \sqrt{\frac{T_0}{\mu}}

Squaring both sides: T33μ=116T0μ  ⟹  T3=316T0  ⟹  T3T0=316(T)\frac{T_3}{3\mu} = \frac{1}{16} \frac{T_0}{\mu} \implies T_3 = \frac{3}{16} T_0 \implies \frac{T_3}{T_0} = \frac{3}{16} \quad \text{(T)}

4. String-4 (IV)

  • Mass per unit length: μ4=4μ\mu_4 = 4\mu
  • Length: L4=7L04L_4 = \frac{7L_0}{4}
  • Harmonic: n4=14n_4 = 14

f4=142(7L04)T44μ=4L0(12T4μ)=2L0T4μf_4 = \frac{14}{2 \left(\frac{7L_0}{4}\right)} \sqrt{\frac{T_4}{4\mu}} = \frac{4}{L_0} \left(\frac{1}{2}\sqrt{\frac{T_4}{\mu}}\right) = \frac{2}{L_0} \sqrt{\frac{T_4}{\mu}}

Since f4=f0f_4 = f_0: 2L0T4μ=12L0T0μ\frac{2}{L_0} \sqrt{\frac{T_4}{\mu}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

T4μ=14T0μ\sqrt{\frac{T_4}{\mu}} = \frac{1}{4} \sqrt{\frac{T_0}{\mu}}

Squaring both sides: T4μ=116T0μ  ⟹  T4=116T0  ⟹  T4T0=116(U)\frac{T_4}{\mu} = \frac{1}{16} \frac{T_0}{\mu} \implies T_4 = \frac{1}{16} T_0 \implies \frac{T_4}{T_0} = \frac{1}{16} \quad \text{(U)}

Conclusion

Matching the strings with List-II:

  • I→P\text{I} \rightarrow \text{P}
  • II→Q\text{II} \rightarrow \text{Q}
  • III→T\text{III} \rightarrow \text{T}
  • IV→U\text{IV} \rightarrow \text{U}

Thus, the correct option is B.

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