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