The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the $x$-projection of the radius vector of the rotating pa

Question

The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the $x$-projection of the radius vector of the rotating particle P in each case.

Motion (a): a circle of radius $A$, period of revolution $T = 4$ s. At $t = 0$ the particle P is at an angle of $45°$ above the positive $x$-axis, and it revolves in the anticlockwise sense.

Motion (b): a circle of radius $B$, period of revolution $T = 30$ s. At $t = 0$ the particle P is at the topmost point of the circle, on the positive $y$-axis, and it revolves in the clockwise sense.

![Figure — two circular motions](/figures/oscillations/two-circular-motions-radius-period-4s-30s-x-projection-fig-13-4-example.svg)

Accepted Answer

> [!TLDR]
> Write the position vector's angle with the $x$-axis as a function of time, then take
> its $x$-projection $x(t) = R\cos(	ext{angle})$. Case (a): $x(t) = A\cos\left(\frac{\pi}{2}t + \frac{\pi}{4}ight)$, SHM of period 4 s. Case (b): $x(t) = B\sin\left(\frac{\pi}{15}tight)$, SHM of period 30 s.

## Why This Approach Works

When a particle P moves uniformly on a circle of radius $R$ with angular speed
$\omega$, its position vector makes an angle $	heta(t)$ with the $+x$-axis that
changes linearly with time. The $x$-projection of P, $x(t) = R\cos	heta(t)$, then
automatically executes SHM — this is exactly the reference-circle connection between
uniform circular motion and SHM developed in this section. The only work needed per
case is to write down $	heta(t)$ correctly from the given initial angle, period, and
sense of rotation (clockwise subtracts from the initial angle; anticlockwise adds to
it).

## Step-by-Step Solution

**Case (a): radius $A$, $T = 4$ s, initial angle $45° = \pi/4$, anticlockwise.**

The angular speed is $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\ 	ext{rad/s}$.

Since the motion is anticlockwise, the angle with the $x$-axis *increases* with time:

$$
	heta(t) = \frac{\pi}{2}t + \frac{\pi}{4}
$$

The $x$-projection is therefore

$$
x(t) = A\cos\left(\frac{\pi}{2}t + \frac{\pi}{4}ight)
$$

This is SHM of amplitude $A$, angular frequency $\pi/2$ rad/s (period 4 s), and
initial phase $\pi/4$.

**Case (b): radius $B$, $T = 30$ s, initial angle $90°=\pi/2$ (on the $+y$-axis), clockwise.**

The angular speed is $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{30} = \dfrac{\pi}{15}\ 	ext{rad/s}$. …

Concept understanding — Simple Harmonic Motion

Simple Harmonic Motion: The Natural Rhythm of Things

The Intuition: A Restoring Force That Fights Displacement

The Precise Statement

What Does This Motion Look Like?

The Mathematical Description (Derived from $F = -kx$ )

Why This Approach Works

Step-by-Step Solution

Why This Approach Works

Step-by-Step Solution

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