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JEE PHYSICS MCQ: LAWS OF MOTION

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One end of a long string of linear mass density 8.0×10−3 kg m−18.0 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} is connected to an electrically driven tuning fork of frequency 256 Hz256 \mathrm{~Hz}, such that the frequency of tuning fork is equal to that of the string. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg90 \mathrm{~kg}. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0t=0, the left end (fork end) of the string x=0x=0 has zero transverse displacement (y=0)(y=0) and is moving along positive yy direction. The amplitude of the wave is 5.0 cm5.0 \mathrm{~cm}. Then the transverse displacement yy (in m) as function of xx and tt that describes the wave on the string. (Take g=9.8 m/s2g=9.8 \mathrm{~m} / \mathrm{s}^2 )
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Correct option: A
0.05sin⁡(1.61×103t−4.84x)0.05 \sin \left(1.61 \times 10^3 \mathrm{t}-4.84 \mathrm{x}\right)
Option A (0.05sin⁡(1.61×103t−4.84x)0.05 \sin \left(1.61 \times 10^3 \mathrm{t}-4.84 \mathrm{x}\right)) is correct because it represents the transverse displacement of the wave with the correct amplitude, angular frequency, and wave number.

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