question_answer

A uniform rope of length L and mass mi hangs vertically from a rigid support. A block of mass \[{{m}_{2}}\]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\[{{\lambda }_{2}}/{{\lambda }_{1}}\] is      

A)  \[\sqrt{\frac{{{m}_{1}}}{{{m}_{2}}}}\]          

B)  \[\sqrt{\frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{2}}}}\]

C) \[\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}}\]           

D) \[\sqrt{\frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}}}\]

Correct Answer: B

Solution :

[b] From figure, tension \[{{T}_{1}}={{m}_{2}}g\] \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}})g\] As we know \[Velocity\propto \sqrt{T}\] So, \[\lambda \propto \sqrt{T}\] \[\Rightarrow \,\,\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{\sqrt{{{T}_{1}}}}{\sqrt{{{T}_{2}}}}\] \[\Rightarrow \,\,\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\sqrt{\frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{2}}}}\]