JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(...\,{m} / {s}.\)
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
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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