Question

# The length of the wire between the pulleys is 1.5 m and its mass is 12 gm. Find the frequency of vibration, with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest. Use value of √11250≈106

A
35.6 Hz
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B
40.3 Hz
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C
70.6 Hz
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D
80.7 Hz
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Solution

## The correct option is A 70.6 Hz The wire between pulley will vibrate in 2 loops. The wire in contact with pulley will behave as fixed point. ⇒ for two loops, n=2 L=n×(λ2) ⇒1.5=2×λ2 ∴λ=1.5 m Mass per unit length μ=mL=12×10−31.5kg/m From equilibrium condition of block. T=mg=90N Velocity of a wave on a stretched string ⇒v=√Tμ=√90×1.512×10−3 ⇒v=√11250≈106 Applying, v=fλ ⇒f=vλ=106λ=1061.5=70.6 Hz

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