A simple harmonic progressive wave in a gas has a particle displacement of y=a at time `t=T//4` at the orgin of the wave and a particle velocity of y =v at the same instant but at a distance `x=lambda//4` from the orgin where T and `lambda` are the periodic time and wavelength of the wave respectively. then for this wave.

Options

A. the amplitude `A` of the wave is `A=2a`
B. the amplitude `A` of the wave is `A=a`
C. the equation of the wave can be represented by
`y asin(v)/(a)[(t)-(x)/(V)]`
D. The equation of the wave can be represented by
`y 2a cos(v)/(a)[t-(x)/(V)]`

Answer

b.,c.

Explanation

Let the equation to the wave be
`y=A sin[2pi((t)/(T)-(x)/(lambda))+phi]`
where `A` is the amplitude of the wave and `phi`, phase angle. It is given that `y=a`, when `x=0` and `t=T//4` and also that `y=v when `x=lambda//4` and `t=T//4`
substituting in (i),
`y=a=A sin((pi)/(2)+phi)`
`y=v=(2piA)/(T)cos[2pi((1)/(4)-(1)/(4) )+phi]`
`v=(2piA)/(T)cosphi`
putting `phi=0, y=a=A`, so that amplitude `A=a`
Also, `v=(2piA)/(T)[cos0]=(2pia)/(T)`
`(2pi)/(T)=(v)/(a)`
Hence the equation to the wave is
`y=a sin(v)/(a)[t-(Tx)/(lambda)]`
` `y=a sin(v)/(a)[t-(x)/(V)]`
where `V=(lambda)/(T)` is the velocity of the wave in the gas.