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A uniform rope of length ' $L$ ' and mass ' $\mathrm{m}_1$ ' hangs vertically from a rigid support. A block of mass ' $\mathrm{m}_2$ ' is attached to the free end of the rope. A transverse wave of wavelength ' $\lambda_1$ ' is produced at the lower end of the rope. The wavelength of the wave when it reaches the top of the rope is ' $\lambda_2$ '. The ratio $\frac{\lambda_1}{\lambda_2}$ is
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The correct answer is:
$\left[\frac{\mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right]^{\frac{1}{2}}$
Let velocity of pulse at lower end be $v_1$ and at top be $\mathrm{v}_2$
$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1} \quad\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{n}} \text { and } \mathrm{n}=\text { constant }\right)$
Velocity of transverse wave on a string is
$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
where, $m$ is linear density.
In this case, $\mathrm{v} \propto \sqrt{\mathrm{T}}$
$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{\left(\mathrm{m}_2+\mathrm{m}_1\right)}{\mathrm{m}_2}}$
Where, $T_2$ is tension at upper end of rope and $T_1$ is tension at lower end of rope.
$\Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2}{m_2+m_1}}$
$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1} \quad\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{n}} \text { and } \mathrm{n}=\text { constant }\right)$
Velocity of transverse wave on a string is
$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
where, $m$ is linear density.
In this case, $\mathrm{v} \propto \sqrt{\mathrm{T}}$
$\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{\left(\mathrm{m}_2+\mathrm{m}_1\right)}{\mathrm{m}_2}}$
Where, $T_2$ is tension at upper end of rope and $T_1$ is tension at lower end of rope.
$\Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2}{m_2+m_1}}$
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