Animated Solution for Physics - Oscillations: A particle executes simple harmonic motion represented by displacement function as $x(t)=A\sin (\omega t+\varphi )$ If the position and velocity of the particle at $t=0$ s are 2 cm and $2\omega$ cm s${}^{-1}$ respectively, then its amplitude is $x\sqrt{2}$ cm, where the value of $x$ is ............ .

A Phasor Approach

Have you ever looked at a swinging pendulum or a bouncing spring and wondered how we can describe its motion so elegantly? The secret lies in the mathematics of Simple Harmonic Motion (SHM). In this problem, we are given the displacement function of a particle executing SHM, and we need to find its amplitude based on its initial position and velocity. Let's dive into the fascinating world of oscillations!

The problem provides us with the general equation for the displacement of a particle executing SHM:

Here, $A$ represents the amplitude (the maximum displacement from the mean position), $\omega$ is the angular frequency (how fast the particle oscillates), and $\varphi$ is the initial phase angle (where the particle starts its journey at $t=0$).

We are given two crucial pieces of information about the particle at the very beginning of its motion ($t=0$): 1. Its position is $2\text{ cm}$. 2. Its velocity is $2\omega {\text{ cm s}}^{-1}$.

Our goal is to find the amplitude $A$ and express it in the form $x\sqrt{2}$ to determine the value of $x$.

Let's start by using the first piece of information. We substitute $t=0$ and $x(0)=2$ into our displacement equation:

This is our first master equation. It tells us that the initial displacement is simply the amplitude multiplied by the sine of the initial phase angle.

Now, what about the velocity? To find the velocity function, we need to differentiate the displacement function with respect to time $t$:

We are given that the initial velocity $v(0)$ is $2\omega$. Let's substitute $t=0$ into our newly found velocity equation:

Notice how $\omega$ appears on both sides of the equation? We can elegantly cancel it out, leaving us with our second master equation:

We now have a system of two equations with two unknowns ($A$ and $\varphi$): 1. $A\sin \varphi =2$ 2. $A\cos \varphi =2$

How do we solve for $A$ without getting tangled up in finding $\varphi$ first? Enter the most famous trigonometric identity: ${\sin }^2\theta +{\cos }^2\theta =1$.

If we square both of our master equations and add them together, something magical happens:

Factoring out $A^2$ on the left side:

Since the term inside the parentheses is exactly $1$, the equation simplifies beautifully:

Taking the square root of both sides, we find the amplitude:

The problem states that the amplitude is of the form $x\sqrt{2}\text{ cm}$. By comparing our result with this form, we can easily find the value of $x$:

Therefore, the value of $x$ is exactly $2$.

A Quick Bonus Insight: What if we wanted to find the initial phase angle $\varphi$? We could simply divide our first master equation by the second:

This tells us that $\varphi =45^{\circ }$ or $\frac{\pi }{4}$ radians. The particle started its motion exactly halfway between the mean position and the extreme position!