Wave Mechanics Question 662
Question: 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
Options:
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}}} $
Show Answer
Answer:
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}}} $
BETA
