A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wavetrain of wavelength 6 cm is produced at the lower end of the rope. What is the wavelength of the wavetrain (in cm) when it reaches the top of the rope ?

Waves PYQ — JEE Main 2020 · 3 Sep · Shift 1 · Q52 with Solution

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Correct answer: A

\mu = \frac{M}{L} = \frac{6}{12} = 0.5\,\text{kg m}^{-1}

At the very bottom, the rope supports only the attached block.

T_{\text{bottom}} = mg = 2g

At the top, the rope supports the block as well as the entire rope.

T_{\text{top}} = (m + M)g = (2+6)g = 8g

For a transverse wave on a string/rope,

v = \sqrt{\frac{T}{\mu}}

So,

\frac{v_{\text{top}}}{v_{\text{bottom}}} = \sqrt{\frac{T_{\text{top}}}{T_{\text{bottom}}}} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2

Thus,

v_{\text{top}} = 2v_{\text{bottom}}

As the wavetrain travels through the rope, its frequency is determined by the source and remains unchanged.

Using

v = f\lambda

with constant $f$ , wavelength is proportional to speed:

\frac{\lambda_{\text{top}}}{\lambda_{\text{bottom}}} = \frac{v_{\text{top}}}{v_{\text{bottom}}} = 2

Therefore,

\lambda_{\text{top}} = 2 \times 6\,\text{cm} = 12\,\text{cm}

Hence, the correct answer is A.