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Question

A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ2λ1 is

A
m1+m2m2
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B
m1m2
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C
m1+m2m1
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D
m2m1
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Solution

The correct option is A m1+m2m2
FBD of the string-block system



At point B, tension in the string T1=m2g

At point A, tension in the string T2=(m1+m2)g

Velocity of a wave in a string is given by

v=Tμ

Mass density of the string μ is not changing.

Hence, ratio of velocities at B and A is

v1v2=T1T2=m2m1+m2

Since v×λ(wavelength)=f(frequency) which is constant for a wave,

λ1λ2=v2v1=m1+m2m2

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