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  • 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

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 \lambda_{1} is produced at the lower end of the rope. the wavelength of the pulse when it reaches the top of the rope is \lambda_{2}. The ratio of \frac{\lambda_{2}}{\lambda_{1}} is:

Option: 1

\sqrt{\frac{m_{2}}{m_{1} + m _{2}}}


Option: 2

\sqrt{\frac{m _{2}}{m_{1}}}


Option: 3

\sqrt{\frac{m _{1}}{m_{2}}}


Option: 4

None of the above


Answers (1)

best_answer

Tension at the lower end of the rope, T1 = m2g

Tension at the top of the rope, T2 = (m1 + m2)g

The velocity of the transverse wave,

        v = n\lambda = \sqrt {\frac{T}{\mu}}

Hence,

            \lambda \alpha \sqrt {T}

 \frac{\lambda_{1}}{\lambda_{2}} = \sqrt {\frac{T_{1}}{T_{2}}} = \sqrt {\frac{m_{2}g}{(m_{1} + m_{2})g}}

= \sqrt {\frac{m_{2}}{m_{1} + m_{2}}}

Posted by

seema garhwal

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