SIMPLE HARMONIC MOTION AND UNIFORM CIRCULAR MOTION -

Consider a particle moving in a circle of radius A with a uniform circular motion with angular speed ω.

Let the particle start at point M where it makes an angle Φ with +ve x-axis. After time t, the particle reaches point N. It is clear from fig when the particle moves in a uniform circular motion, its projection on y-axis execute in S.H.M between its two extreme position A and B with amplitude A

from fig. $\fn_cm \large \sin (\omega t+\phi)=\frac{y}{A}$

$\fn_cm \large \left [ y=A\sin(\omega t+\phi) \right ]$

this is the general equation of S.H.M where

y – Displacement of a particle at time t from the mean position

A – Amplitude (maximum displacement of the particle)

ω – Angular frequency

t – Time taken by the particle

Φ – Phase constant

ωt+Φ – Phase

NOTE

From fig, projection of particle on x-axis also execute in S.H.M with the general equation $\fn_cm \large x=A\cos(\omega t+\phi)$ which is out of phase $\fn_cm \large \frac{\pi}{2}$ from $\fn_cm \large y=\sin(\omega t+\phi)$
In spite of this connection between circular motion and SHM, the force acting on a particle in linear SHM is very different from the centripetal force needed to keep a particle in a uniform circular motion.

For example, suppose a particle moves in a uniform circular motion of radius R . The initial position and the sense of rotation are indicated as shown in the figure.

from fig (a) after time t, P makes an angle $\fn_cm \large \omega t+ \frac{\pi}{4}\;\;=\;\;\frac{2\pi}{T}+\frac{\pi}{4}$ with the x-axis

The projection of P on the x-axis is given by

$\fn_cm \large x=R\cos(\frac{2\pi}{T}t+\frac{\pi}{4})$

for T=4 sec

$\fn_cm \large x=R\cos(\frac{2\pi}{4}t+\frac{\pi}{4})$

$\fn_cm \large x=R\cos(\frac{\pi}{2}t+\frac{\pi}{4})$ $\fn_cm \large x=R\cos(\frac{\pi}{2}+\frac{\pi}{4})$

which is the SHM of amplitude R, period 4s and an initial phase= $\fn_cm \large \frac{\pi}{4}$

from fig (b) after time t, P makes an angle $\fn_cm \large \frac{\pi}{2}-\omega t\;\;=\;\;\frac{\pi}{2}-\frac{2\pi}{T}t$ with the x-axis

The projection of P on the x-axis is given by

$\fn_cm \large x=R\cos(\frac{\pi}{2}-\frac{2\pi}{T}t)$

$\fn_cm \large x=R\sin(\frac{2\pi}{T}t)$

for T=30 sec

$\fn_cm \large x=R\sin(\frac{\pi}{15}t)$

we can write here

$\fn_cm \large x=R\cos(\frac{\pi}{15}t-\frac{\pi}{2})$

which is the SHM of amplitude R, period 30 sec and an initial phase= $\fn_cm \large -\frac{\pi}{2}$

a. 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.

b. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x = –2 sin (3t + π/3)
(b) x = cos (π/6 – t)
(c) x = 3 sin (2πt + π/4)
(d) x = 2 cos πt