A particle executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 m/s 2. The maximum velocity of the particle is (in m/s )

Options

A. 0.01
B. 0.05
C. 0.5
D. 0.25

Answer

To solve the problem, we need to find the maximum velocity of a particle executing simple harmonic motion (SHM) given its amplitude, displacement, and acceleration at a certain instant.

**Step-by-step solution:**

1. **Identify the given values:**
- Amplitude (A) = 0.1 m
- Displacement (x) = 0.02 m
- Acceleration (a) = 0.5 m/s²

2. **Use the relationship between acceleration and displacement in SHM:**
In simple harmonic motion, the acceleration (a) is related to the displacement (x) by the formula:
\[ a = -\omega^2 x \]
For our calculations, we can ignore the negative sign since we are interested in magnitudes:
\[ a = \omega^2 x \]

3. **Rearranging the formula to find angular frequency (ω):**
From the equation \( a = \omega^2 x \), we can express ω as:
\[ \omega = \sqrt{\frac{a}{x}} \]

4. **Substituting the known values to find ω:**
\[ \omega = \sqrt{\frac{0.5 \, \text{m/s}^2}{0.02 \, \text{m}}} \]
\[ \omega = \sqrt{25} = 5 \, \text{rad/s} \]

5. **Calculate the maximum velocity (V₀):**
The maximum velocity (V₀) in SHM is given by the formula:
\[ V_0 = A \cdot \omega \]
Substituting the values we have:
\[ V_0 = 0.1 \, \text{m} \cdot 5 \, \text{rad/s} = 0.5 \, \text{m/s} \]

6. **Final answer:**
The maximum velocity of the particle is \( 0.5 \, \text{m/s} \).

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