Question

The acceleration of a particle executing simple harmonic motion at a distance of 10 cm from the mean position is 0.4 ms\(^{-2}\). If the amplitude of the particle is 30 cm, the maximum acceleration of the particle is

• A. 0.6 ms\(^{-2}\)
• B. 0.9 ms\(^{-2}\)
• C. 1.2 ms\(^{-2}\)
• D. 1.5 ms\(^{-2}\)

Hint:

In SHM, use \( a = \omega^2 x \) to find \( \omega \), then calculate the maximum acceleration using \( a_{\text{max}} = \omega^2 A \). Ensure units are consistent (convert cm to m).

Answer 1

The Correct Option is C

In simple harmonic motion (SHM), the acceleration \( a \) at a displacement \( x \) from the mean position is given by: \[ a = \omega^2 x \] where \( \omega \) is the angular frequency. Given:
- \( x = 10 \, \text{cm} = 0.1 \, \text{m} \),
- \( a = 0.4 \, \text{ms}^{-2} \).
Substituting the values: \[ 0.4 = \omega^2 \times 0.1 \quad \Rightarrow \quad \omega^2 = \frac{0.4}{0.1} = 4 \, \text{s}^{-2} \] The maximum acceleration in SHM occurs at the amplitude \( A \), and is given by: \[ a_{\text{max}} = \omega^2 A \] Given the amplitude \( A = 30 \, \text{cm} = 0.3 \, \text{m} \), substitute \( \omega^2 = 4 \): \[ a_{\text{max}} = 4 \times 0.3 = 1.2 \, \text{ms}^{-2} \] So, the maximum acceleration of the particle is 1.2 ms\(^{-2}\).

Answer 2

Step 1: Understand the given data
- Displacement from mean position, \( x = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Acceleration at this point, \( a = 0.4 \, \text{m/s}^2 \)
- Amplitude, \( A = 30 \, \text{cm} = 0.3 \, \text{m} \)

Step 2: Use the formula for acceleration in simple harmonic motion
\[ a = \omega^2 x \]
where \( \omega \) is the angular frequency.

Step 3: Calculate angular frequency \( \omega \)
\[ \omega^2 = \frac{a}{x} = \frac{0.4}{0.1} = 4 \] \[ \Rightarrow \omega = 2 \, \text{rad/s} \]

Step 4: Calculate maximum acceleration \( a_{\max} \)
\[ a_{\max} = \omega^2 A = 4 \times 0.3 = 1.2 \, \text{m/s}^2 \]

Final answer:
The maximum acceleration of the particle is \( 1.2 \, \text{m/s}^2 \).