Question

A uniform rope of length $10\text{m}$ and mass $15\text{kg}$ hangs vertically from a rigid support. A block of mass $5\text{kg}$ is attached to the free end of the rope. A transverse pulse of wavelength $0.08\text{m}$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope will be
$($Take $g=10{\text{m/s}}^2)$

• A. $0.36\text{m}$
• B. $0.26\text{m}$
• C. $0.16\text{m}$
• D. $0.06\text{m}$

Answer 1

The correct option is C $0.16\text{m}$

From FBD of block,

Applying equilibrium condition, tension in the rope at lower end,
$T_1=50\text{N}$
Similarly, tension in the rope at the top end,
$T_2=50+150=200\text{N}$


Velocity of wave pulse,
$v=\sqrt{\frac{T}{\mu }}$
For the same rope,
$v\propto \sqrt{T}$
So, $\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$ ..........$\left(1\right)$
Also, $v=\lambda f$
For the same source,
$v\propto \lambda$
So, $\frac{v_1}{v_2}=\frac{{\lambda }_1}{{\lambda }_2}$ ..........$\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_1}{T_2}}$
$\Rightarrow \frac{0.08}{{\lambda }_2}=\sqrt{\frac{50}{200}}$
$\Rightarrow {\lambda }_2=0.16\text{m}$

Why this question? Tips: Once the wave pulse is generated then the associated momentum and energy changes due to interaction of the wave with particles of the medium. However, frequency remains unchanged.