A transverse wave is represented by the equation
`y=y_0sin.(2pi)/(lamda)(vt-x)`
For what value of `lamda`, the maximum particle velocity equal to two times the wave velocity?

Options

A. `pi A`
B. `pi^(2)A`
C. `(pi A)/(2)`
D. `pi A^(2)`

Answer

A

Explanation

`y = A cos 2pi (n t - (x)/(lamda)) " but " n = (v)/(lamda)`
`:. y = A cos 2pi ((vt)/(lamda) - (x)/(lamda))`
`= A "cos" (2pi)/(lamda) (vt - x)`
`:.` The particle velocity is given by
`((dy)/(dt)) = - A (2pi v)/(lamda) "sin"(2pi)/(lamda) (vt - x)`
This is maximum when `"sin" (2pi)/(lamda) (vt - x) = 1`
`:. ((dy)/(dt))_("max") = A (2pi v)/(lamda)` (numerically)
`:. A .(2pi v)/(lamda) = 2v " (given)" :. lamda = pi A`