JEE Main | 2018Basic Maths and Units and DimensionsHard

Question

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 ×103 kg/m3 and its Young's modulus is 9.27×1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

Options

A.

2.5 kHz

B.

10 kHz

C.

7.5 kHz

D.

5 kHz

Solution

Velocity of wave = Yρ
=9.27×10102.7×103
=3.433×107
= 103 ×34.33
vω=5.85×103m/sec
Since rod is clamped at middle fundamental wave shape is as follow


λ2=L
λ= 2L
L = 60 cm = 0.6 m (given)
λ  = 1.2 m
v = fλ
f=vλ=5.85×1031.2
= 4.88 ×103 Hz ; 5 KHz

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