Exploring Simple Harmonic Motion: Finding the Amplitude

This question delves into the characteristics of Simple Harmonic Motion (SHM). We are given information about the speed of a particle at different displacements and asked to find the oscillation's amplitude. Let's break down how to solve this using the fundamental equations of SHM.

Understanding the SHM Speed-Displacement Relation

In Simple Harmonic Motion, the speed ($v$) of a particle is related to its displacement ($x$) from the mean position, the amplitude ($A$), and the angular frequency ($\omega$) by the following equation:

$v = \omega \sqrt{A^2 - x^2}$

Here:

• $v$ is the instantaneous speed of the particle.
• $\omega$ is the constant angular frequency of the motion.
• $A$ is the amplitude (maximum displacement) of the motion.
• $x$ is the instantaneous displacement from the mean position.

The angular frequency ($\omega$) depends on the system's properties (like mass and spring constant) and remains the same throughout the motion. Our goal is to find the amplitude ($A$).

Step-by-Step Calculation of Amplitude

We are given two scenarios:

1. When the displacement is $x$, the speed is $v$.
2. When the displacement is $2x$, the speed is $v/2$.

Let's apply the SHM speed-displacement formula to both scenarios:

Using the formula $v = \omega \sqrt{A^2 - x^2}$ for the first case (displacement $x$, speed $v$), we get:

$v = \omega \sqrt{A^2 - x^2} \quad \quad (1)$

For the second case (displacement $2x$, speed $v/2$), the formula becomes:

$\frac{v}{2} = \omega \sqrt{A^2 - (2x)^2}$

$\frac{v}{2} = \omega \sqrt{A^2 - 4x^2} \quad \quad (2)$

To simplify calculations and eliminate the square roots, let's square both Equation (1) and Equation (2):

From (1): $v^2 = \omega^2 (A^2 - x^2)$

From (2): $\left(\frac{v}{2}\right)^2 = \omega^2 (A^2 - 4x^2)$

$\frac{v^2}{4} = \omega^2 (A^2 - 4x^2)$

Since $\omega$ is constant for this motion, we can eliminate it by dividing the squared equation from Step 1 by the squared equation from Step 2:

$\frac{v^2}{\frac{v^2}{4}} = \frac{\omega^2 (A^2 - x^2)}{\omega^2 (A^2 - 4x^2)}$

Simplifying the left side gives 4. The $\omega^2$ terms cancel out:

$4 = \frac{A^2 - x^2}{A^2 - 4x^2}$

Now, we rearrange the equation to solve for $A$:

$4(A^2 - 4x^2) = A^2 - x^2$

Distribute the 4:

$4A^2 - 16x^2 = A^2 - x^2$

Group the $A^2$ terms on one side and the $x^2$ terms on the other:

$4A^2 - A^2 = 16x^2 - x^2$

$3A^2 = 15x^2$

Isolate $A^2$:

$A^2 = \frac{15x^2}{3}$

$A^2 = 5x^2$

Take the square root of both sides to find $A$. Since amplitude represents a distance, it must be positive:

$A = \sqrt{5x^2}$

$A = x\sqrt{5}$

Final Amplitude Result

By applying the SHM speed-displacement relationship and solving the resulting equations, we find that the amplitude of the oscillation is $A = x\sqrt{5}$. This matches the third option provided.