JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 24)
The amplitude of wave disturbance propagating in the positive x-direction is given by $$y = {1 \over {{{(1 + x)}^2}}}$$ at time t = 0 and $$y = {1 \over {1 + {{(x - 2)}^2}}}$$ at t = 1 s, where x and y are in metres. The shape of wave does not change during the propagation. The velocity of the wave will be ___________ m/s.
Answer
2
Explanation
As per question,
at t = 0, y = $${1 \over {{{(1 + x)}^2}}}$$
and at t = 1 s, y = $${1 \over {1 + {{(x - 2)}^2}}}$$ .... (i)
As we know,
At t = t s, y = $${1 \over {1 + {{(x - vt)}^2}}}$$
So, at t = 1 s, y = $${1 \over {1 + {{(x - v)}^2}}}$$ .... (ii)
On comparing Eqs. (i) and (ii), we get
v = 2 ms$$-$$1
Hence, the velocity of the wave will be 2 m/s.
at t = 0, y = $${1 \over {{{(1 + x)}^2}}}$$
and at t = 1 s, y = $${1 \over {1 + {{(x - 2)}^2}}}$$ .... (i)
As we know,
At t = t s, y = $${1 \over {1 + {{(x - vt)}^2}}}$$
So, at t = 1 s, y = $${1 \over {1 + {{(x - v)}^2}}}$$ .... (ii)
On comparing Eqs. (i) and (ii), we get
v = 2 ms$$-$$1
Hence, the velocity of the wave will be 2 m/s.
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