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Find the equation for simple harmonic mo...

Find the equation for 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.

A

`x=0.04 "sin"(200pit+(pi)/(3))`

B

`x=0.04 "sin"(100pit+(pi)/(3))`

C

`x=0.04 "sin"(100pit+(pi)/(4))`

D

`x=0.04 "sin"(100pit+(pi)/(6))`

Text Solution

Verified by Experts

The correct Answer is:
B
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))`
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