JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2021
- Medium
- Numerical
The amplitude of wave disturbance propagating in the positive \(x\)-direction is given by \(y=\frac{1}{(1+x)^{2}}\) at time \(t=0\) and \(y=\frac{1}{1+(x-2)^{2}}\) 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 \(...\,{m} / {s}.\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
At \({t}=0, {Y}=\frac{1}{1+{x}^{2}}\)
\(\text { At time } t=t, y=\frac{1}{1+(x-v t)^{2}}\)
\(\text { At } t=1, y=\frac{1}{1+(x-v)^{2}}....(i)\)
\(\text { At } t=1, y=\frac{1}{1+(x-2)^{2}}....(ii)\)
comparing \((i) \& (ii)\)
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