Question

A uniform rope of length $12 \,m$ and having a mass $6\, kg$ hangs vertically from a rigid support. A block of mass $2\, kg$ is attached to the free end of rope. A transverse pulse of wavelength $0.06\, m$ 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

Tension at the top $=(6+2) \times 10=80 \,N$
Tension at the bottom $=2 \times 10=20 \,N$
$\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.
$\therefore \frac{v_{\text {top }}}{v_{\text {bottom }}}=\frac{\lambda_{\text {top }}}{\lambda_{\text {bottom }}}=\sqrt{\frac{80}{20}}=2$
$\therefore \lambda_{\text {top }}=(2 \times 0.06)=0.12\, m =a$
$\therefore 5 \alpha=0.6 \,m$