#### Question

Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure

Obtain the corresponding simple harmonic motions of the *x*-projection of the radius vector of the revolving particle P, in each case.

#### Solution 1

1) Let A be any point on the circle of reference of the figure (a) From A, draw BN perpendicular on x axis

if `angle POA = theta`, then

`angleOAM = theta = omegat`

:. In triangle OAM,

`(OM)/(OA) = sintheta`

`:. (-x)/3 = omegat = sin (2pi)/T t`

`:. x = -3 sin (2pi)/2 t or x = -3sin pit` which is the equation of SHM.

2) Let B be any point on the circle of reference of figure (b). From B draw BN perpendicular on x-axis

Then `triangleBON = theta = omegat`

:. In `triangleONB`, cos theta = `(ON)/(OB)`

or `ON = OB cos theta`

`:. - x = 2 cos omega t`

`=> x =- 2 cos (2pi)/T t = -2 cos (2pi)/4 t`

`:. x = - 2 cos pi/4 t` which is equation of SHM

#### Solution 2

a) Time period, *T *= 2 s

Amplitude, *A* = 3 cm

At time, *t *= 0, the radius vector OP makes an angle `pi/2` with the positive *x-*axis, i.e., phase angle `phi = + pi/2`

Therefore, the equation of simple harmonic motion for the* x*-projection of OP, at time* t*, is given by the displacement equation:

`x = A cos[(2pit)/T + phi]`

`= 3 cos ((2pit)/2 + pi/2) = -3sin ((2pit)/2)`

`:. x = - 3 sinpit " cm"`

**(b) **Time period, *T *= 4 s

Amplitude, *a* = 2 m

At time *t* = 0, OP makes an angle π with the *x*-axis, in the anticlockwise direction. Hence, phase angle, *Φ* = + π

Therefore, the equation of simple harmonic motion for the *x*-projection of OP, at time *t*, is given as:

`x = acos ((2pit)/T + phi ) = 2 cos ((2pit)/4 + pi)`

`:. x = - 2 cos (pi/2 t) m`