Question:
The acceleration of a particle performing SHM is \( 12 \, \text{cm/s}^2 \) at a distance of 3 cm from the mean position. Its time period is:
The acceleration of a particle performing SHM is \( 12 \, \text{cm/s}^2 \) at a distance of 3 cm from the mean position. Its time period is:
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In SHM, the acceleration is directly proportional to the displacement. Using the relationship between acceleration, displacement, and angular frequency, we can calculate the time period.
Updated On: Apr 22, 2026
- 0.5 s
- 1.0 s
- 2.0 s
- 3.14 s
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The Correct Option is D
Solution and Explanation
Step 1: Acceleration in SHM.
In simple harmonic motion (SHM), the acceleration \( a \) is related to the displacement \( x \) from the mean position by the equation: \[ a = - \omega^2 x \] where \( \omega \) is the angular frequency and \( x \) is the displacement.
Step 2: Use the given data.
We are given:
- \( a = 12 \, \text{cm/s}^2 = 0.12 \, \text{m/s}^2 \)
- \( x = 3 \, \text{cm} = 0.03 \, \text{m} \) Substitute these values into the SHM equation: \[ 0.12 = \omega^2 \times 0.03 \]
Step 3: Solve for angular frequency \( \omega \).
Rearranging the equation: \[ \omega^2 = \frac{0.12}{0.03} = 4 \] \[ \omega = 2 \, \text{rad/s} \]
Step 4: Calculate the time period.
The time period \( T \) of the particle is related to the angular frequency by the formula: \[ T = \frac{2\pi}{\omega} \] Substitute \( \omega = 2 \, \text{rad/s} \): \[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \] Thus, the time period is approximately \( 3.14 \, \text{seconds} \).
Step 5: Conclusion.
Therefore, the time period is \( 3.14 \, \text{seconds} \).
In simple harmonic motion (SHM), the acceleration \( a \) is related to the displacement \( x \) from the mean position by the equation: \[ a = - \omega^2 x \] where \( \omega \) is the angular frequency and \( x \) is the displacement.
Step 2: Use the given data.
We are given:
- \( a = 12 \, \text{cm/s}^2 = 0.12 \, \text{m/s}^2 \)
- \( x = 3 \, \text{cm} = 0.03 \, \text{m} \) Substitute these values into the SHM equation: \[ 0.12 = \omega^2 \times 0.03 \]
Step 3: Solve for angular frequency \( \omega \).
Rearranging the equation: \[ \omega^2 = \frac{0.12}{0.03} = 4 \] \[ \omega = 2 \, \text{rad/s} \]
Step 4: Calculate the time period.
The time period \( T \) of the particle is related to the angular frequency by the formula: \[ T = \frac{2\pi}{\omega} \] Substitute \( \omega = 2 \, \text{rad/s} \): \[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \] Thus, the time period is approximately \( 3.14 \, \text{seconds} \).
Step 5: Conclusion.
Therefore, the time period is \( 3.14 \, \text{seconds} \).
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