Class 11 > Unit # 11:Oscillations > Uniform Circular Motion and S.H.M.

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Equations of Displacement, Velocity, Acceleration & Time Period of SHM

Video Lecture: Uniform Circular Motion & SHM Derivations

Watch the complete step-by-step video lecture explaining the mathematical equations of displacement, velocity, acceleration, and time period:

1. Equation of Displacement

At some instant of time $t$, the angle between radius vector $OP$ and the x-axis is given by $(\omega t + \phi)$, where $\phi$ is the initial phase angle (the angle which $OP$ makes with the x-axis at time $t = 0$).

From the right-angled triangle $OPQ$:

$\sin(\omega t + \phi) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{x}{x_0}$

$x = x_0 \sin(\omega t + \phi)$

Where $x_0$ represents the amplitude of S.H.M. of projection $Q$, and $x$ is the instantaneous displacement.

2. Equation of Acceleration

A particle $P$ moving along the circumference of a circle of radius $r$ (or $x_0$) with linear velocity $V_p$ has an angular velocity $\omega$ given by:

$\omega = \frac{V_p}{r}$ or $V_p = r\omega$

As particle $P$ moves along the circular path, its projection $Q$ executes vibratory motion along $AOC$. The centripetal acceleration $a_c$ of particle $P$ is directed towards the center:

$a_c = \frac{V_p^2}{r} = r\omega^2$

The component of centripetal acceleration along the line of motion of $Q$ is:

$a_o = a_c \cos\theta = r\omega^2 \cos\theta$

From the geometry of the triangle, since $r \cos\theta = x$:

$a_o = -\omega^2 x$

The negative sign indicates that the acceleration of $Q$ is directed towards the center (mean position) and is directly proportional to displacement $x$, which is the defining characteristic of Simple Harmonic Motion.

3. Equation of Velocity

The velocity of projection $Q$ ($V_Q$) is equal to the x-component of the velocity of particle $P$ ($V_p$) directed along $AOC$:

$V_Q = V_p \sin\theta = x_0 \omega \sin\theta \quad \text{--- (i)}$

Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, we get $\sin\theta = \sqrt{1 - \cos^2\theta}$. Since $\cos\theta = \frac{x}{x_0}$, substituting this into the expression gives:

$V_Q = x_0 \omega \sqrt{1 - \frac{x^2}{x_0^2}} = \omega \sqrt{x_0^2 - x^2} \quad \text{--- (ii)}$

Special Cases for Velocity:

• At Extreme Positions ($x = \pm x_0$):
  $V_Q = \omega \sqrt{x_0^2 - x_0^2} = 0$
  The velocity of projection at the extreme position is zero.
• At Mean Position ($x = 0$):
  $V_Q = \omega \sqrt{x_0^2 - 0} = x_0 \omega$ (Maximum)
  The velocity of projection at the mean position is maximum.

4. Time Period

The time required to complete one full cycle of motion is called the time period, denoted by "$T$".

According to the definition of angular velocity:

$\omega = \frac{\theta}{t}$

For one complete cycle, angular displacement $\theta = 2\pi$ and time $t = T$:

$\omega = \frac{2\pi}{T} \implies T = \frac{2\pi}{\omega}$

Fig: Geometric parameters for displacement, velocity, and acceleration derivations in SHM.

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