Correct Answer - maximum amplitude `20 units ` wave velocity is `5000 units`. Frequency `100 units` loop length is `25 units`
We know the equation of a stationary wave is given by
`y = A sin [ ( 2 pi)/( lambda) ( vt - x)] + A sin [ ( 2 pi)/(lambda) ( vt + x)]`
`= 2 A cos ( 2pi x)/( lambda) sin ( 2pi v t)/( lambda)`
` = R sin ( 2pi)/( lambda) vt`
Here , `R = 2 A cos (2 pi x// lambda)` is the amplitude of medium particle situated at a distance `x`.
The given equation can be expressed as
`y = 10 sin [( 2 pi)/( 50) (5000 t + x)]`
` = 2 xx 10 cos (( 2pi x)/( 50)) sin (( 2pi)/(50) 5000 t)`
Comparing it with standard equation of stationary wave , we get wavelength `lambda = 50 units`
Wave velocity ` v = 5000 units`
Thus amplitude `R = 2 xx 10 cos ( 2 pi x)/( 50)`
and maximum amplitude `R_(max) = 2 xx 10 = 20 units`
Frequency `= (v)/( lambda) = ( 5000)/(50) = 100 units`
And loop length is
`(lambda)/(2) = (50)/(2) = 25 units`
