AP EAMCET · PHYSICS · Oscillations
If the displacement of a particle executing simple harmonic motion is given by \(\mathrm{x}=0.5 \cos (125.6 \mathrm{t})\), then the time period of oscillation of the particle is nearly
(Here x is displacement in metre and t is time in second)
- A 1 s
- B 2 s
- C 0.09 s
- D 0.05 s
Answer & Solution
Correct Answer
(D) 0.05 s
Step-by-step Solution
Detailed explanation
\(\omega = 125.6 \text{ rad/s}\) \(T = \frac{2\pi}{\omega}\) \(T = \frac{2 \times 3.14159}{125.6} = 0.05 \text{ s}\)
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