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The amplitude of a wave disturbance propagating along positive X-axis is given by `=1/(1+x^(2))` at t=0 and `y=1/[1+(x-2)^(2)]` at t=4 s where x and y are in metre. The shape of wave diturbance does not change with time. The velocity of the wave isA. `0.5 m//s`B. `1 m//s`C. `2 cm/s`D. `4 m//s` |
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Answer» Correct Answer - a `y=(1)/(1+(x^(2))` at `t=0` and `y=(1)/(1+(x-2)^(2))` at `t=4s` `v=(Deltax)/(Deltat)=(x-(x-2))/(4-0)=(2)/(4)=0.5 m//s` |
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