A particle vibrates harmonically at a frequency of 0.5 Hz. At the initial moment it is in an equilibrium position moving at a speed of 20 cm/s. Write down the equation of the vibrations.

Answer

`s=0.064cos(pit+(3pi)/(2))=0.064sinpit`.

Explanation

The law of oscillations is of the form `s=Acos(omegat+varphi)`. Since `s_(0)=0`, it follows that `0=Acosvarphi`, giving `varphi=(2k+1)(pi)/(2)` The initial plase is always less than the period of the oscillations `varphilt2pi`. Hence `varphi=pi//2`, or `varphi=3pi//2`.
The particle.s velocity is `v=-Aomegasin(omegat+varphi)`. The initial velocity `v_(0)=-Aomegasinvarphi=0.20` is, according to the statement to the problem, a positive quantity, and this is possible only if `varphi=3pi//2`. Hence, `Aomega=0.2`. But `omega=2piv=pi` rad/s. Hence the amplitude is `A=0.20//pi=0.064m`. Knowing the amplitude, the frequency and the initial phase we may easily wirte down the equation of the oscillations.