The equation of a transverse wave travelling in a rope is given by `y = 5 sin (4t - 0.02 x)`, where y and x are in cm and time t is in second. Then the maximum transverse speed of wave in the rope is

The equation of a transverse wave travelling on a rope is given by y =10 sin pi (0.01xx -2.00t) where y and x are in cm and t is in seconds. The maximum transverse speed of a particle in the rope is about

The equation of a transverse wave travelling on a rope is given by y = 10sin pi (0.01x-2.000t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about

The equation of a transverse wave travelling in a rope is given by y=7 sin (4.0 t-0.02 x) where y and x are in cm and the time is in second. Calculate (i) the maximum transverse speed and (ii) the maximum particle acceleration?

The equation of a transverse travelling on a rope is given by y=10sinpi (0.01x-2.00t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about

The equation of a transverse travelling on a rope is given by y=10sinpi (0.01x-2.00t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about

The equation of a transverse wave travelling on a rope given by y=10sinpi(0.01x-2.00t) whrer y and x are in cm and t in second .This maximum traverse speed of a particle in the rope is about

The equation of a transverse wave travelling on a rope given by y=10sinpi(0.01x-2.00t) whrer y and x are in cm and t in second .This maximum traverse speed of a particle in the rope is about

The equation of a transverse wave is given by y = 10 sin {pi (0.01 x - 2t)} where y and x are in cm and t is in seconds. Its frequency is ………..