Question

A particle performing linear S.H.M. of amplitude $0.1\text{ m}$ has displacement $0.02\text{ m}$ and acceleration $0.5\text{ m/s}^2$. The maximum velocity of the particle in m/s is

• A. $0.05$
• B. $0.50$
• C. $0.01$
• D. $0.25$

Hint:

When computing $\omega^2 = \frac{a}{x}$, clear the decimals immediately by shifting the numerator and denominator by two decimal places: $\frac{0.50}{0.02} = \frac{50}{2} = 25$. Recognizing that 25 is a perfect square lets you extract $\omega = 5$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem provides information about a particle undergoing linear Simple Harmonic Motion (S.H.M.).
We are given its amplitude ($A = 0.1\text{ m}$), an instantaneous displacement ($x = 0.02\text{ m}$), and the magnitude of its acceleration at that displacement ($a = 0.5\text{ m/s}^2$).
We need to determine the maximum velocity ($v_{max}$) that this particle achieves during its path.

Step 2: Key Formula or Approach:
1. The magnitude of acceleration in linear S.H.M. as a function of displacement is given by:
$$a = \omega^2 x$$ 2. The maximum velocity of a particle in S.H.M. occurs at the mean position and is given by:
$$v_{max} = A\omega$$ Where $\omega$ represents the angular frequency of the oscillation.

Step 3: Detailed Explanation:
First, use the acceleration formula to find the value of $\omega^2$:
$$\omega^2 = \frac{a}{x}$$ Substitute the given values into this relation:
$$\omega^2 = \frac{0.5}{0.02} = \frac{50}{2} = 25$$ Taking the square root on both sides gives the angular frequency:
$$\omega = \sqrt{25} = 5\text{ rad/s}$$ Now, calculate the maximum velocity ($v_{max}$) using the calculated angular frequency and the given amplitude ($A = 0.1\text{ m}$):
$$v_{max} = A\omega = 0.1 \times 5 = 0.5\text{ m/s}$$

Step 4: Final Answer:
The maximum velocity of the particle is $0.50\text{ m/s}$, which corresponds exactly to option (B).