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Question

A 50 cm long wire having a mass of 20 g is fixed at two ends and is vibrated in its fundamental mode. A 75 cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its first overtone mode by the vibrating wire. Find the tension in the wire. Speed of sound in air is 340 m/s.

A
2446 N
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B
4264 N
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C
4624 N
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D
4552 N
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Solution

The correct option is C 4624 N Given that Length of wire, lw=50 cm=50×10−2 m Mass of wire, m=20 g=20×10−3 kg Linear density of the wire, μ=mlw=20×10−350×10−2=250=125 kg/m So fundamental frequency of the wire is fw=f1=12lw√Tμ=12×50×10−2  ⎷T(125)−−−−−(1) Length of closed organ pipe, lP=75 cm=75×10−2 m Speed of sound, v=340 m/s Resonance frequency of closed organ pipe is given by fn=(2n−1)v4lP for n=1,2,3.. First overtone of closed organ pipe is fP=3v4lP=3×3404×75×10−2−−−−−(2) Since both wire and organ pipe are in resorance, fw=fP ⇒12×50×10−2  ⎷T(125)=3×3404×75×10−2 ⇒√T=3×3404×75×10−2×2×50×10−25 ⇒√T=68 ⇒T=4624 N

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