Answer to Question #80636 in Physics for Madhu

Question

An object does simple harmonic vibration along X axis. Its amplitude is 5.0*10^-2m,f=2.0hz.when t=0,vibration moves along X axis passing through equilibrium. Find expression of vibration

Answer 1

The general law of harmonic vibration
x(t)=x_max sin⁡(2πft+φ)
Where
x_max is an amplitude,
f is a frequency
φ is an initial phase
So
x(t)=5.0×〖10〗^(-2) sin⁡〖(4πt+φ) (m)〗
Since x(0)=0, we get sin⁡〖φ=0,φ=0〗
Finally
x(t)=5.0×〖10〗^(-2) sin⁡〖(4πt) (m)〗