# 17.3 Uniform Circular Motion and SHM This section explores the connection between Uniform Circular Motion (UCM) and Simple Harmonic Motion (SHM). It demonstrates how the projection of an object undergoing UCM onto a diameter of its circular path exhibits SHM. This relationship allows for the derivation of key SHM equations for displacement, velocity, and acceleration. The projection of a point P, moving with constant angular velocity $\omega$ in a circle, onto any diameter of that circle executes Simple Harmonic Motion. - **Reference Circle**: The circular path followed by point P is called the reference circle. - **Projection**: The point Q, which is the perpendicular projection of P onto a diameter (such as the x-axis), oscillates back and forth along this diameter. - **Relationship**: The period of revolution for P is the same as the period of oscillation for Q. The angular speed $\omega$ of P is equivalent to the angular frequency of Q's SHM. The radius of the circle, $x_0$, corresponds to the amplitude of the SHM. ### Expressions for Displacement, Velocity, and Acceleration By analyzing the motion of point P in the reference circle, we can derive the kinematic equations for its projection Q, which is undergoing SHM. #### 1. Displacement The position of P at any time t is given by the angle $\theta = \omega t$. The displacement of the projection Q along the x-axis is the adjacent side of the triangle OPQ. **Formula**: $x = x_0 \cos(\omega t)$ #### 2. Velocity The velocity of P ($v_p$) is tangential to the circle with magnitude $v_p = x_0 \omega$. The velocity of Q ($v$) is the horizontal component of $v_p$. **Formula**: $v = \omega \sqrt{x_0^2 - x^2}$ #### 3. Acceleration The acceleration of P ($a_p$) is the centripetal acceleration, directed towards the center O, with magnitude $a_p = x_0 \omega^2$. The acceleration of Q ($a$) is the horizontal component of $a_p$. The negative sign indicates it is a restoring acceleration, always directed towards the equilibrium position O. **Formula**: $a = -\omega^2 x$ ### Phase The state of motion of a particle executing SHM is described by its phase. In the context of UCM, the phase angle $\theta = \omega t + \phi$ determines the displacement and direction of motion. ## Worked Examples **Example 17.2**: A body with mass 800 g vibrates with amplitude 30 cm. The restoring force is 60 N at displacement 0.30 m. Find (a) spring constant $k$, (b) angular frequency $\omega$, (c) velocity and (d) acceleration at $x = 12$ cm. **Solution**: **(a) Spring constant**: Using Hooke's Law, $F = kx_0$. $$k = \frac{F}{x_0} = \frac{60\ \mathrm{N}}{0.30\ \mathrm{m}} = 200\ \mathrm{N\ m}^{-1}$$ **(b) Angular frequency**: $$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.8}} = 15.81\ \mathrm{rad\ s}^{-1}$$ **(c) Velocity at $x = 12$ cm**: $$v = \omega \sqrt{x_0^2 - x^2} = 15.81 \times \sqrt{(0.30)^2 - (0.12)^2}$$ $$v = 15.81 \times \sqrt{0.09 - 0.0144} = 15.81 \times \sqrt{0.0756}$$ $$v = 15.81 \times 0.275 = 4.35\ \mathrm{m\ s}^{-1}$$ **(d) Acceleration at $x = 12$ cm**: $$a = -\omega^2 x = -(15.81)^2 \times 0.12$$ $$a = -249.9 \times 0.12 = -29.99\ \mathrm{m\ s}^{-2}$$ | Quantity | Formula | Description | | :--- | :--- | :--- | | **Displacement** | $x = x_0 \cos(\omega t)$ | Position at time t | | **Velocity** | $v = \omega \sqrt{x_0^2 - x^2}$ | Velocity at displacement x; maximum at $x = 0$, zero at $x = \pm x_0$ | | **Acceleration** | $a = -\omega^2 x$ | Acceleration at displacement x; maximum at $x = \pm x_0$, zero at $x = 0$ | This model is a key tool for visualizing and analyzing oscillatory systems, from simple pendulums to molecular vibrations. The projection method connects UCM directly to SHM equations, enabling analysis of velocity and acceleration in oscillatory systems.