JEE Advance - Physics (2014 - Paper 1 Offline - No. 10)
One end of a taut string of length 3 m along the x-axis is fixed at x = 0. The speed of the waves in the string is 100 ms$$-$$1. The other end of the string is vibrating in the y-direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary wave is (are)
$$y(t) = A\sin {{\pi x} \over 6}\cos {{50\pi t} \over 3}$$
$$y(t) = A\sin {{\pi x} \over 3}\cos {{100\pi t} \over 3}$$
$$y(t) = A\sin {{5\pi x} \over 6}\cos {{250\pi t} \over 3}$$
$$y(t) = A\sin {{5\pi x} \over 2}\cos 250\pi t$$
Explanation
The displacement of a stationary wave is given by
$$y(x,t) = A\sin \left( {{{2\pi x} \over \lambda }} \right)\cos (2\pi vt)$$.
The boundary conditions give node at x = 0 and anti-node at x = 3 m i.e.,
y(0, t) = 0, ....... (1)
y(3, t) = $$\pm$$A. ..... (2)
The fundamental frequency is given by
$${v_0} = {v \over \lambda } = {v \over {4l}} = {{100} \over {4(3)}} = {{25} \over 3}$$ Hz.
Thus, the waveform will satisfy equations (1) and (2), and the permissible frequencies will be odd multiples of v0.
$$y(x,t) = A\sin \left( {{{2\pi x} \over \lambda }} \right)\cos (2\pi vt)$$.
The boundary conditions give node at x = 0 and anti-node at x = 3 m i.e.,
y(0, t) = 0, ....... (1)
y(3, t) = $$\pm$$A. ..... (2)
The fundamental frequency is given by
$${v_0} = {v \over \lambda } = {v \over {4l}} = {{100} \over {4(3)}} = {{25} \over 3}$$ Hz.
Thus, the waveform will satisfy equations (1) and (2), and the permissible frequencies will be odd multiples of v0.
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