Q.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).

Concept: Simple Harmonic Motion — each SHM can be represented as the projection of uniform circular motion on a diameter. The standard form is $x = A\cos(\omega t + \phi)$, where $A$ is the radius, $\omega$ the angular speed, and $\phi$ the initial phase.

Step 1 — Rewrite each in cosine form

1. $x = -2\sin(3t + \pi/3) = 2\cos(3t + \pi/3 + \pi/2) = 2\cos(3t + 5\pi/6)$
2. $x = \cos(\pi/6 - t) = \cos(t - \pi/6)$
3. $x = 3\sin(2\pi t + \pi/4) = 3\cos(2\pi t + \pi/4 - \pi/2) = 3\cos(2\pi t - \pi/4)$
4. $x = 2\cos\pi t$ (already in standard form) Step 2 — Extract parameters

The Concept: Why a Circle?

Simple harmonic motion is the projection of uniform circular motion onto a diameter. Imagine a particle moving anticlockwise on a circle of radius $A$ with constant angular speed $\omega$. At time $t$, its angular displacement from the positive $x$-axis is $\omega t + \phi$. The $x$-coordinate of this particle is $A\cos(\omega t + \phi)$ — that's our SHM.

So every SHM of the form $x = A\cos(\omega t + \phi)$ corresponds to:

• Radius = $A$ (the amplitude)
• Angular speed = $\omega$
• Initial phase = $\phi$ (the angle at $t=0$ from the positive $x$-axis)
• Initial position = $A\cos\phi$

The trick is to get every given equation into that exact cosine form. Sine can be converted using $\sin\theta = \cos(\theta - \pi/2)$.

(a) $x = -2\sin(3t + \pi/3)$

Step 1: Convert to cosine form.

First, handle the minus sign: $-2\sin\theta = 2\sin(\theta + \pi)$ because $\sin(\theta + \pi) = -\sin\theta$.

So $x = 2\sin(3t + \pi/3 + \pi) = 2\sin(3t + 4\pi/3)$.

Now convert sine to cosine: $\sin\alpha = \cos(\alpha - \pi/2)$.

Thus $x = 2\cos(3t + 4\pi/3 - \pi/2) = 2\cos(3t + 5\pi/6)$.

Step 2: Read off parameters.

• Amplitude $A = 2$ cm → radius of circle = 2 cm
• Angular speed $\omega = 3$ rad/s
• Initial phase $\phi = 5\pi/6$ rad (150°)

Step 3: Initial position.

$x_0 = A\cos\phi = 2\cos(5\pi/6) = 2(-\sqrt{3}/2) = -\sqrt{3}$ cm ≈ -1.73 cm.

(b) $x = \cos(\pi/6 - t)$

Step 1: Rewrite as $\cos(\omega t + \phi)$.

Cosine is even: $\cos(\theta) = \cos(-\theta)$. So $\cos(\pi/6 - t) = \cos(t - \pi/6)$.

Now it's in standard form: $x = 1\cdot\cos(1\cdot t - \pi/6)$.

Step 2: Read off parameters.

• $A = 1$ cm
• $\omega = 1$ rad/s
• $\phi = -\pi/6$ rad (-30°)

Step 3: Initial position.

$x_0 = \cos(-\pi/6) = \cos(\pi/6) = \sqrt{3}/2$ cm ≈ 0.866 cm.

(c) $x = 3\sin(2\pi t + \pi/4)$

Step 1: Convert sine to cosine.

$\sin\theta = \cos(\theta - \pi/2)$.

So $x = 3\cos(2\pi t + \pi/4 - \pi/2) = 3\cos(2\pi t - \pi/4)$.

Step 2: Read off parameters.

• $A = 3$ cm
• $\omega = 2\pi$ rad/s
• $\phi = -\pi/4$ rad (-45°)

Step 3: Initial position.

$x_0 = 3\cos(-\pi/4) = 3(\sqrt{2}/2) = \frac{3\sqrt{2}}{2}$ cm ≈ 2.12 cm. …

Step 1: Convert every given expression to the standard reference-circle form $x=A\cos(\omega t+\phi)$, using $\sin\theta=\cos(\theta-\pi/2)$ and $-\sin\theta=\cos(\theta+\pi/2)$.

Step 2 — (a): $x=-2\sin(3t+\pi/3)=2\cos(3t+\pi/3+\pi/2)=2\cos(3t+5\pi/6)$, so $A=2\ \text{cm}$, $\omega=3\ \text{rad/s}$, $\phi=5\pi/6$.

Step 3 — (b): $x=\cos(\pi/6-t)=\cos(t-\pi/6)$ (cosine is even), so $A=1\ \text{cm}$, $\omega=1\ \text{rad/s}$, $\phi=-\pi/6$.

Step 4 — (c): $x=3\sin(2\pi t+\pi/4)=3\cos(2\pi t+\pi/4-\pi/2)=3\cos(2\pi t-\pi/4)$, so $A=3\ \text{cm}$, $\omega=2\pi\ \text{rad/s}$, $\phi=-\pi/4$. …