Question

The equation of a travelling wave is given by
y=+(b)/(a)sqrt(a^(2)-(x-ct)^(2)) where -altxlta`
=0,otherwise
Find the amplitude and the wave velocity fot the wave, what is the initial particle velocity at the position `x=a//2`?

Answer 1

The amplitude of wave velocity is given by the maximum particle displacement `y_(max)` which, evidently, occurs when
`(x-ct)^(2)=0`, i.e., `y(max)=amplitude=b//asqrt(a^(2)=b)`.
the combination `(x-ct)`, appearing in the wave equation, when compared with the expression `x-vt`, reveals that, the wave velocity is `v=c`.
Different the displacement equation with respect to time `t`, we get,
`(dy)/(dt)=(b)/(2a)1/sqrt(a^(2)-(x-ct)^(2))[-2(x-ct)](-c)`
`=(bc(x-ct))/(asqrt(a^(2)-(x-ct)^(2)`
Putting `x=a//2` and `t=0,` we get initial particle velocity
`(dy)/(dt)=(bc(a//2))/(asqrta^(2)-(a^(2)//4))=(bcsqrt3)/(3a)`