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: Get $\omega$ from the acceleration-displacement link.
In SHM, acceleration magnitude is $a = \omega^2 x$. The single data pair $(x, a)$ pins down $\omega$ without needing energy methods.
Step 2: Solve for $\omega^2$.
$\omega^2 = \dfrac{a}{x} = \dfrac{0.5}{0.02} = 25\,\text{s}^{-2}$.
Step 3: Find $\omega$.
$\omega = \sqrt{25} = 5\,\text{rad/s}$.
Step 4: Recall the max-velocity formula.
Maximum speed occurs at the mean position: $v_{\max} = A\omega$.
Step 5: Substitute amplitude and $\omega$.
$v_{\max} = 0.1 \times 5 = 0.5\,\text{m/s}$.
Step 6: Conclude.
The maximum velocity is $0.50\,\text{m/s}$. \[ \boxed{v_{\max} = 0.50\ \text{m/s}} \]