AP EAMCET · PHYSICS · Oscillations
The kinetic energy of a particle executing simple harmonic motion at a displacement of 3 cm from the mean position is 4 mJ. If the amplitude of the particle is 5 cm, then the maximum force acting on the particle is
- A 0.25 N
- B 0.50 N
- C 0.75 N
- D 1.25 N
Answer & Solution
Correct Answer
(A) 0.25 N
Step-by-step Solution
Detailed explanation
\( K = \frac{1}{2} m \omega^2 (A^2 - x^2) \) \( 4 \times 10^{-3} = \frac{1}{2} m \omega^2 ((0.05)^2 - (0.03)^2) \) \( m \omega^2 = \frac{2 \times 4 \times 10^{-3}}{(0.05)^2 - (0.03)^2} = \frac{8 \times 10^{-3}}{0.0016} = 5 \, \text{N/m} \) \( F_{max} = m \omega^2 A \)…
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