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Question:

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 L₀ and 2L₀. It is found that in string -1 (μ) at free length L₀ and tension T₀ the fundamental mode frequency is f₀. List - I List - II (I) String - 1 (μ) (P) 1 (II) String - 2(2μ) (Q) 1/2 (III) String - 3 (3μ) (R) √2 (IV) String - 4(4μ) (S) √3 (T) 3/16 (U) 1/16 The length of the strings 1, 2, 3 and 4 are kept fixed at L₀, (3L₀)/2, (5L₀)/4 and (7L₀)/4, respectively. Strings 1, 2, 3 and 4 are vibrated at their 1st, 3rd, 5th and 14th harmonics, respectively such that all the strings have the same frequency. The correct match for the tension in the four strings in the units of T₀ will be :

I→P,II→Q,III→R,IV→T

I→P,II→R,III→T,IV→U

I→T,II→Q,III→R,IV→U

I→P,II→Q,III→T,IV→U

Solution:

Correct option is C. I→P,II→Q,III→T,IV→U Case 1. L=L₀, T=T₀, f=f₀ f₁=1/(2L₀)√(T₀/μ) Case 2. L=(3L₀)/2 f₂=(3/2)×(1/(3L₀/2))√(T₂/(2μ))=f₀ ⇒f₀=1/(2L₀)√(T₂/(2μ)) ⇒T₂=T₀/2 Case 3. L=(5L₀)/4 f₃=(5/2)×(1/(5L₀/4))√(T₃/(3μ))=f₀ ⇒f₀=2/(3L₀)√(T₃/(3μ)) ⇒T₃=(3T₀)/16 case 4. L=(7L₀)/4 ⇒f₄=(14/2)×(1/(7L₀/4))√(T₄/(4μ))=f₀ ⇒f₀=2/(L₀)√(T₄/(4μ)) ⇒T₄=T₀/16