When 0.98 m long metallic wire is stretched, an extension of 0.02 m is produced. An organ pipe 0.5 m long & open at both ends, when sounded with this stressed metallic wire, produces 8 beats in its fundamental mode of both the instruments. By decreasing the strain in the wire, the number of beats are found to decrease. Find Young's modulus of the wire. The density of metallic wire is 10 4 kgm -3 & sound velocity in air is 292 ms -1 .
Text Solution
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(Y = 1.76 x 10 11 N/m 2 )
The fundamental frequency of open pipe is n = 
= 
= 292 Hz
Let L be the initial length and L + Δ L
the stressed length of the wire then L + Δ L = .98 + .02 = 1 m
The fundamental frequency of stressed wire n = 
where r is the radius and d is the density of wire.
The stressed wire and the organ pipe (frequency 292 Hz) produces 8 beats
∴ n = 292 
8
By decreasing tension velocity decreases At the same time number of beats also decreases.
∴ n = 292 + 8 = 300
By eqn. we have 
= 4 n 2 d (L + Δ L) 2
Y = 
= 
= 
= 17.64 × 10 10 N/m 2
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