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Question

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.330 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at the fundamental frequency. Then:
[Take speed of sound in air as 343 m/s]

A
Fundamental frequency of air column is 71.45 Hz.
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B
Fundamental frequency of air column is 35 Hz.
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C
The tension in the wire is 29.5 N.
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D
The tension in the wire is 64.50 N.
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Solution

The correct option is D The tension in the wire is 64.50 N.
Given that,
Length of tube (L)=1.20 m
Length of wire (l)=0.330 m
Mass of wire (m)=9.60 g
Speed of sound (vs)=343 m/s

Frequency of closed organ pipe is given by
fn=(2n1)v4l where n=1,2,3,4....
Fundamental frequency, f=v4Lf=343 4×1.20=71.45 Hz

Now, linear mass density of wire,
μ=ml=9.60×1030.33=0.029 kg/m

Frequency of wave on the wire is given by f=n2lTμ
Fundamental frequency of wave on the wire
f=12lTμ
T=(f×2l)2×μ
Substituting the given data, we get
T=(71.45×2×0.33)2×0.029
[since frequency of oscillation of wire is the same as the fundamental frequency of air column]
T=64.4864.50 N
Hence, options (a) and (d) are correct.

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