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 (integer)
2
Solution
As per question, <br/><br/>at t = 0, y = ${1 \over {{{(1 + x)}^2}}}$<br/><br/>and at t = 1 s, y = ${1 \over {1 + {{(x - 2)}^2}}}$ .... (i)<br/><br/>As we know,<br/><br/>At t = t s, y = ${1 \over {1 + {{(x - vt)}^2}}}$<br/><br/>So, at t = 1 s, y = ${1 \over {1 + {{(x - v)}^2}}}$ .... (ii)<br/><br/>On comparing Eqs. (i) and (ii), we get<br/><br/>v = 2 ms<sup>$-$1</sup><br/><br/>Hence, the velocity of the wave will be 2 m/s.
About this question
Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types
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