A travelling wave has the frequency `upsilon` and the particle displacement amplitude `A`. For the wave the particle velocity amplitude is ………………… and the particle acceleration amplitude is ……………………..

Answer

`A(2piv),A(2piv)^(2)`

Explanation

If `y=A sin (omegat-kx),`displacement amplitude=A (Maximum displacement)
Particle velocity,`v=(dy)/(dt)=A omega cos(omegat-kx)`
Therefore,velocity amplitude `=Aomega=A(2piv)` Particle acceleration, `A"cc"=(dv)/(dt)=-Aomega^(2) sin (omegat-kx)`
Therefore,acceleration (maximum acceleration) amplitude `=Aomega^(2)=A(2piv)^(2)`