InterviewSolution
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InterviewSolution
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A wave moving with constant speed on a uniform string passes the point `x=0` with amplitude `A_0`, angular frequency `omega_0` and average rate of energy transfer `P_0`. As the wave travels down the string it gradually loses energy and at the point `x=l`, the average rate of energy transfer becomes. At the point `x=l`. Angular frequency and amplitude are respectively:A. `omega_(0)` and` (A_(0))/(sqrt2)`B. `(omega_0)/(sqrt2)` and `A_0`C. less that `omega_(0)` and `A_(0)`D. `(omega_0)/(sqrt2)` and `(A_0)/(sqrt2)` |
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Answer» Correct Answer - A `P=(1)/(2)muomega^2A^2V` and `v=sqrt((T)/(mu))` will not change as both T and `mu` are constant. `omega` will also not change as it is property of the source only that is causing the wave motion. Hence to make power half the amplitude Becomes `(A_0)/(sqrt2)` |
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