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PhyMedAdv 2019 • Paper 2

Highest Fundamental Frequency Matching for Vibrating Strings

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}

Assuming the tension in each string is T0T_0, select the correct matching for the maximum attainable fundamental frequency measured in units of f0f_0.

Options

A

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

B

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

Correct
C

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

D

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

Topics & Concepts

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Step-by-Step Solution

To find the maximum attainable fundamental frequency for each string, we start with the general expression for the fundamental frequency ff of a stretched string of length LL, tension TT, and mass per unit length μ\mu:

f=12LTμf = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

For String-1 with linear mass density μ\mu, at free length L=L0L = L_0 and tension T=T0T = T_0, the fundamental frequency is given as f0f_0:

f0=12L0T0μf_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}

Since the tension in each string is fixed at T=T0T = T_0, and the length LL can be adjusted within the range L0≤L≤2L0L_0 \le L \le 2L_0, the fundamental frequency for any string ii (with mass per unit length μi\mu_i) is maximized when the length LL is minimized, i.e., at L=L0L = L_0:

fi,max=12L0T0μif_{i, \text{max}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu_i}}

Now, we evaluate fi,maxf_{i, \text{max}} for each string in terms of f0f_0:

  1. For String-1 (μ1=μ\mu_1 = \mu): f1,max=12L0T0μ=1⋅f0f_{1, \text{max}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} = 1 \cdot f_0 Therefore, (I) →\rightarrow (P).

  2. For String-2 (μ2=2μ\mu_2 = 2\mu): f2,max=12L0T02μ=12(12L0T0μ)=12f0f_{2, \text{max}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{2\mu}} = \frac{1}{\sqrt{2}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{\sqrt{2}} f_0 Therefore, (II) →\rightarrow (R).

  3. For String-3 (μ3=3μ\mu_3 = 3\mu): f3,max=12L0T03μ=13(12L0T0μ)=13f0f_{3, \text{max}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{3\mu}} = \frac{1}{\sqrt{3}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{\sqrt{3}} f_0 Therefore, (III) →\rightarrow (S).

  4. For String-4 (μ4=4μ\mu_4 = 4\mu): f4,max=12L0T04μ=12(12L0T0μ)=12f0f_{4, \text{max}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{4\mu}} = \frac{1}{2} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{1}{2} f_0 Therefore, (IV) →\rightarrow (Q).

Comparing with the options given in the question: I→P,II→R,III→S,IV→Q\text{I} \rightarrow \text{P}, \quad \text{II} \rightarrow \text{R}, \quad \text{III} \rightarrow \text{S}, \quad \text{IV} \rightarrow \text{Q}

This corresponds to Option (B).

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