Question

An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with frequency of $(\frac{25}{\pi}) Hz$. At the position $x = 0.04 m$, the object has kinetic energy 1 J and potential energy 0.6 J. The amplitude of oscillation is

• A. 0.06 m
• B. 0.6 m
• C. 0.08 m
• D. 0.8 m

Hint:

Total energy in SHM remains constant and is equal to potential energy at the extreme position.

Answer 1

The Correct Option is C

Step 1: Total Energy
$E = K.E. + P.E. = 1 + 0.6 = 1.6 \text{ J}$
Step 2: Total Energy Formula
$E = \frac{1}{2} m \omega^2 A^2$
Step 3: Calculation of Angular Velocity
$\omega = 2 \pi f = 2 \pi (\frac{25}{\pi}) = 50 \text{ rad/s}$
Step 4: Solving for Amplitude
$1.6 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2$
$1.6 = 0.1 \times 2500 \times A^2 \Rightarrow 1.6 = 250 A^2$
$A^2 = \frac{1.6}{250} = \frac{16}{2500} \Rightarrow A = \frac{4}{50} = 0.08 \text{ m}$
Final Answer:(C)