The displacement of a particle is given by $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. If this represents a simple harmonic motion, the amplitude of its oscillation is:

Question

Prepp Team · Accepted Answer

Finding the Amplitude in Simple Harmonic Motion

The question asks for the amplitude of oscillation for a particle described by the displacement equation: $y(t) = K + P \sin^2(\omega t) + Q \sin(\omega t) \cos(\omega t)$. A key characteristic of simple harmonic motion (SHM) is that its displacement from the equilibrium position can be represented by a single sine or cosine function, possibly shifted by a phase angle and an equilibrium offset. The general form is often written as $A \sin(\omega t + \phi) + C$, where $A$ is the amplitude.

Simplifying the Displacement Equation using Trigonometric Identities

To identify the amplitude, we need to simplify the given equation into the standard SHM form. We can use fundamental trigonometric identities:

The identity for $\sin^2(\theta)$: $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$
The identity for $\sin(\theta)\cos(\theta)$: $\sin(\theta) \cos(\theta) = \frac{\sin(2\theta)}{2}$

Applying these identities with $\theta = \omega t$ to the terms in the given displacement equation:

$P \sin^2(\omega t) = P \left( \frac{1 - \cos(2\omega t)}{2} \right)$
$Q \sin(\omega t) \cos(\omega t) = Q \left( \frac{\sin(2\omega t)}{2} \right)$

Rewriting the Displacement Equation

Substitute these simplified terms back into the original equation for $y(t)$: $y(t) = K + P \left( \frac{1 - \cos(2\omega t)}{2} \right) + Q \left( \frac{\sin(2\omega t)}{2} \right)$

Now, expand and rearrange the terms to group constants and oscillating components:

$y(t) = K + \frac{P}{2} - \frac{P}{2} \cos(2\omega t) + \frac{Q}{2} \sin(2\omega t)$

Combine the constant terms ($K$ and $\frac{P}{2}$) and factor out $\frac{1}{2}$ from the oscillating terms:

$y(t) = \left( K + \frac{P}{2} \right) + \frac{1}{2} \left( Q \sin(2\omega t) - P \cos(2\omega t) \right)$

Determining the Amplitude from the Simplified Form

The expression $\left( K + \frac{P}{2} \right)$ represents the constant offset or the equilibrium position of the motion.

The term $\frac{1}{2} \left( Q \sin(2\omega t) - P \cos(2\omega t) \right)$ represents the oscillating part. Any expression of the form $a \sin(x) + b \cos(x)$ can be written as a single trigonometric function $R \sin(x + \phi)$ or $R \cos(x + \phi)$, where the amplitude $R = \sqrt{a^2 + b^2}$.

In our case, the oscillating part is $\frac{1}{2} \left( Q \sin(2\omega t) - P \cos(2\omega t) \right)$. Let's focus on the term inside the parenthesis: $Q \sin(2\omega t) - P \cos(2\omega t)$.

This can be written in the form $A' \sin(2\omega t + \phi)$ or $A' \cos(2\omega t + \phi)$. To find the amplitude $A'$ of this combined term, we use the formula $A' = \sqrt{(\text{coefficient of } \sin)^2 + (\text{coefficient of } \cos)^2}$.

Here, the coefficient of $\sin(2\omega t)$ is $Q$, and the coefficient of $\cos(2\omega t)$ is $-P$. So, the amplitude $A'$ is:

$A' = \sqrt{(Q)^2 + (-P)^2}$ $A' = \sqrt{Q^2 + P^2}$

Calculating the Final Amplitude

The oscillating part of the displacement $y(t)$ is $\frac{1}{2} \times (A' \times \text{some phase}) = \frac{1}{2} \times \sqrt{P^2 + Q^2} \times (\text{some sinusoidal function})$.

Therefore, the amplitude of the simple harmonic motion is the coefficient multiplying the sinusoidal function:

Amplitude $= \frac{1}{2} A'$ Amplitude $= \frac{1}{2} \sqrt{P^2 + Q^2}$

Conclusion on SHM Amplitude

By simplifying the given displacement equation using double-angle trigonometric identities, we transformed it into a standard form representing simple harmonic motion. The constant term indicates the equilibrium position, and the coefficient of the resulting sine (or cosine) function represents the amplitude of the oscillation. The calculation confirms that the amplitude is $\frac{1}{2}\sqrt{P^2 + Q^2}$.