Question

The time period of a particle executing simple harmonic motion is $2\pi$ second. If the maximum acceleration of the particle is $10 ms^{-2}$, then the maximum velocity of the particle is

• A. $10 ms^{-1}$
• B. $20 ms^{-1}$
• C. $5 ms^{-1}$
• D. $15 ms^{-1}$

Hint:

Remember the relationship: $a_{max} = \omega \cdot v_{max}$. This allows you to switch between maximum acceleration and velocity without finding the amplitude explicitly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a particle in Simple Harmonic Motion (SHM), the kinematic quantities are related through the angular frequency $\omega$.

• Maximum Velocity: $v_{max} = A\omega$
• Maximum Acceleration: $a_{max} = A\omega^2$

Where $A$ is the amplitude.
Step 2: Calculate Angular Frequency ($\omega$):
Given Time Period $T = 2\pi$ s. \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 \, \text{rad/s} \]
Step 3: Calculate Maximum Velocity:
We can express $v_{max}$ in terms of $a_{max}$ and $\omega$: \[ \frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega \] \[ v_{max} = \frac{a_{max}}{\omega} \] Given $a_{max} = 10 \, ms^{-2}$: \[ v_{max} = \frac{10}{1} = 10 \, ms^{-1} \]
Step 4: Final Answer:
The maximum velocity is $10 \, ms^{-1}$.