Question

If the equation representing the relation between displacement 'x' and velocity 'v' of a particle executing simple harmonic motion is $4v^2 = 25 - x^2$, then the time period of the particle is (All the quantities in the equation are in SI units)

• A. $\pi \text{ s}$
• B. $2\pi \text{ s}$
• C. $4\pi \text{ s}$
• D. $6\pi \text{ s}$

Hint:

For SHM, the fundamental relationship $v = \omega \sqrt{A^2 - x^2}$ is key. Always rearrange the given velocity-displacement equation to match this standard form to easily extract the angular frequency ($\omega$) and amplitude ($A$).

Answer 1

The Correct Option is C

Step 1: Recall the standard equation for velocity in Simple Harmonic Motion (SHM).
For a particle executing Simple Harmonic Motion, the relationship between its velocity ($v$), displacement ($x$), and amplitude ($A$) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where $\omega$ is the angular frequency of oscillation.
Step 2: Rearrange the given equation to match the standard form.
The given equation is:
\[ 4v^2 = 25 - x^2 \]
Divide by 4 to isolate $v^2$:
\[ v^2 = \frac{25 - x^2}{4} \]
\[ v^2 = \frac{1}{4}(25 - x^2) \]
Take the square root of both sides:
\[ v = \sqrt{\frac{1}{4}(25 - x^2)} \]
\[ v = \frac{1}{2}\sqrt{25 - x^2} \]
Step 3: Compare with the standard SHM velocity equation to find $\omega$ and $A$.
Comparing $v = \frac{1}{2}\sqrt{25 - x^2}$ with $v = \omega \sqrt{A^2 - x^2}$:
We can directly identify the angular frequency ($\omega$) and the square of the amplitude ($A^2$):
\[ \omega = \frac{1}{2} \text{ rad/s} \] And \[ A^2 = 25 \quad \Rightarrow \quad A = 5 \text{ m} \] (The amplitude A = 5 m, but we only need $\omega$ for the time period). Step 4: Calculate the time period ($T$).
The time period ($T$) of a particle executing SHM is related to its angular frequency ($\omega$) by the formula: \[ T = \frac{2\pi}{\omega} \] Substitute the value of $\omega$: \[ T = \frac{2\pi}{\frac{1}{2}} \] \[ T = 2\pi \times 2 \] \[ T = 4\pi \text{ s} \] The final answer is $\boxed{4\pi \text{ s}}$.