Question 390: Waves

Question

A uniform rope of length '$$L$$' and mass '$$m_1$$' hangs vertically from a rigid support. A block of mass '$$m_2$$' 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

Accepted Answer

Let velocity of pulse at lower end be $$v_1$$ and at top be $$v_2$$
$$	herefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1} \quad\left(\because \lambda=\frac{\mathrm{v}}{\mathrm{n}} 	ext { and } \mathrm{n}=	ext { constant }ight)$$
Velocity of transverse wave on a string is
$$v=\sqrt{\frac{T}{m}}$$
where, $$\mathrm{m}$$ is linear density.
In this case, $$\mathrm{v} \propto \sqrt{\mathrm{T}}$$
$$	herefore \quad \frac{\lambda_2}{\lambda_1}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{\left(\mathrm{m}_2+\mathrm{m}_1ight)}{\mathrm{m}_2}}$$
Where, $$T_2$$ is tension at upper end of rope and $$T_1$$ is tension at lower end of rope.
$$\Rightarrow \frac{\lambda_1}{\lambda_2}=\sqrt{\frac{m_2}{m_2+m_1}}$$

Answer

$$\left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$$

Answer

$$\left[\frac{m_1+m_2}{m_2}\right]^{\frac{1}{2}}$$

Answer

$$\left[\frac{\mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right]^{\frac{1}{2}}$$

Answer

$$\left[\frac{\mathrm{m}_2}{\mathrm{~m}_1-\mathrm{m}_2}\right]^{\frac{1}{2}}$$

Waves

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