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 :

Options

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)`

Answer

A

Explanation

`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))`