Q.

A string is under tension so that its length is increased by 1ntimes its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be :

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a

1 : n

b

n : 1

c

n : 1

d

n : 1.

answer is C.

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Detailed Solution

Velocity of longitudinal wave v1 = Yρ and velocity of transverse wave V2 = Tm = Tρsv1V2 = YsT = YT/s = YYΔll = nNow, f ∝ v∴ f1f2 = V1V2 = nIn the above equation ρ = density of string, s = area of cross–section of string, Y = Young’s modulus of elasticity.Ts = Stress = Y(strain) = YΔll
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A string is under tension so that its length is increased by 1ntimes its original length. The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be :