Question

A particle executes SHM with amplitude \( 0.1 \text{ m} \) and angular frequency \( 10 \text{ rad/s} \). Maximum acceleration is:

• A. \( 1 \text{ m/s}^2 \)
• B. \( 5 \text{ m/s}^2 \)
• C. \( 10 \text{ m/s}^2 \)
• D. \( 100 \text{ m/s}^2 \)

Hint:

Keep your position boundaries clear in SHM! The velocity is maximum at the mean position (\(x=0\)) where acceleration is zero. Conversely, acceleration reaches its peak magnitude (\(\omega^2 A\)) at the extreme endpoints (\(x = \pm A\)) where the velocity drops to zero!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Acceleration in SHM is a restoring acceleration directed toward the mean position.
Its magnitude increases as the particle moves away from the mean position, reaching a peak at the extreme points (the amplitude).
Key Formula or Approach:
The acceleration \(a\) as a function of displacement \(x\) is:
\[ a = -\omega^2 x \]
Maximum magnitude occurs when \(|x| = A\):
\[ a_{\text{max}} = \omega^2 A \]
Step 2: Detailed Explanation:
Extracting the given variables:
Amplitude \(A = 0.1 \text{ m}\)
Angular frequency \(\omega = 10 \text{ rad/s}\)
Substitute these into the maximum acceleration formula:
\[ a_{\text{max}} = (10)^2 \cdot 0.1 \]
\[ a_{\text{max}} = 100 \cdot 0.1 = 10 \text{ m/s}^2 \]
Note that while the direction of acceleration is always opposite to displacement, the question asks for the "maximum acceleration," which typically refers to the magnitude of this peak value.
Step 3: Final Answer:
The maximum acceleration of the particle is \(10 \text{ m/s}^2\).