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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, same of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector the rotating particle P is

A

`x(t)=B " sin"((2pit)/(30))`

B

`x(t)=B " cos"((pit)/(15))`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
(a) Let angular velocity of the particle executing circular motion is `omega` and when it is at Q makes and angle `theta` as shown in the diagram.
Clearly , `theta=omegat`
Now, we can write OR
`=OQ "cos" (90-theta)=OQ "sin" theta=OQ "sin"omegat`
`=r"sin"omegat" "[because OQ=r]`

`impliesx=r"sin"omegat=B"sin"omegat" "[because r=B]`
`=B"sin"(2pi)/(T)t=B"sin"((2pi)/(30)t)`
Clearly, this equation represents SHM.
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