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Figure shows the circular motion of a pa...

Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is..........

A

`x (t) = B sin ((2pi t)/(30))`

B

`x (t) = B cos ((pi t)/(15))`

C

`x (t) = B sin ((pi t)/(15)+(pi)/(2))`

D

`x (t) = B cos ((pi t)/(15)+(pi)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
Suppose a particle is moving uniformly on a circle of radius r with angular speed `omega` and when it is on point Q it makes an angle `theta` as shown in figure.
Here `theta = omega t`
In right triangle `triangleORQ`
`sin theta = (OR)/(OQ)`
`OR = OQ sin theta`
`=r sin omega t" [Radius of circle OQ = r and " theta = omega t]`
OR is the X-component
`therefore x(t) = B sin ((2pi)/(T)t)" "[therefore r= B, omega = (2pi)/(T)]`
`therefore x(t) = B sin ((2pi)/(30)t)`.
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