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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`P_0`/2. At the point `x=l`. Angular frequency and amplitude are respectively:A. `omega_(0)` and `A_(0)//sqrt(2)`B. `omega_(0)//sqrt(2)` and `A_(0)`C. less than `omega_(0)` and `A_(0)`D. `omega_(0)//sqrt(2)` and `A_(0)//sqrt(2)` |
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Answer» Correct Answer - A `p_(0) = A_(0^(2)) omega_(0^(2)) muv` `(p_(0))/(2) = A^(2)omega^(2)muv` `:. 2 = (A_(0^(2)) omega_(0^(2)))/(A^(2)omega^(2))` `:.` As, `omega = omega_(0)` (frequency remains then same) `:. A = (A_(0))/(sqrt(2))` |
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