Question

The position and velocity of a particle executing simple harmonic motion at $t = 0$ are given by $3$ cm and $8$ cm/s respectively. If the angular frequency of the particle is $2$ rad/s then the amplitude of oscillation, in centimeters, is

• A. $3$
• B. $4$
• C. $5$
• D. $6$
• E. $8$

Hint:

Use $A^2 = x^2 + (v/\omega)^2$ when position and velocity are known at the same instant.

Answer 1

The Correct Option is C

Concept:
In SHM: \[ A^2 = x^2 + \left(\frac{v}{\omega}\right)^2 \]

Step 1: Substitute values.
\[ x = 3,\quad v = 8,\quad \omega = 2 \] \[ A^2 = 3^2 + \left(\frac{8}{2}\right)^2 = 9 + 16 = 25 \]

Step 2: Find amplitude.
\[ A = 5 \text{ cm} \]