Question

A uniform rope of length $12m$ and having a mass $6kg$ hangs vertically from a rigid support. A block of mass $2kg$ is attached to the free end of rope. A transverse pulse of wavelength $0.06m$ is produced at the lower end of rope. Its wavelength when it reaches the top end of the rope is given by $\alpha m$. Find $50\alpha$.

Answer 1

Answer: 6

Tension at the top $=(6+2)\times 10=80N$
Tension at the bottom $=2\times 10=20N$
$\therefore$ Velocity of wave at top $=\sqrt{\frac{T_{\text{top }}}{\mu }}=\sqrt{\frac{80}{\mu }}$.
Velocity of wave at bottom $=\sqrt{\frac{T_{\text{bottom }}}{\mu }}=\sqrt{\frac{20}{\mu }}$
Since source is same $\Rightarrow f$ is same.
$\therefore \frac{v_{\text{top }}}{v_{\text{bottom }}}=\frac{{\lambda }_{\text{top }}}{{\lambda }_{\text{bottom }}}=\sqrt{\frac{80}{20}}=2$
$\therefore {\lambda }_{\text{top }}=(2\times 0.06)=0.12m=a$
$\therefore 5\alpha =0.6m$