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.

Answer 1

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`

Answer 2

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