Question

Find the equation fo simple harmonic motion of a particle whose amplitude is 0.04 and whose frequency is 50 Hz . The initial phase is `pi//3` . Assume that motion of particle is started from mean position.

Answer 1

From equation of SHM , `x=A "sin"(omegat+phi)`
Here, `A=0.04 m,v=50 Hz , phi=(pi)/(3)`
`therefore x=0.04 "sin"(2pi xx 50t+(pi)/(3)) " "(because omega=2piv)`
or `x=0.04 "sin"(100pit+(pi)/(3))`