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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
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`
Solution 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
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