AP EAMCET · PHYSICS · Oscillations
The relation between the force ( F is newton) acting on a particle executing simple harmonic motion and the displacement of the particle ( \(y\) in metre) is \(500 \mathrm{~F}+\pi^2 y=0\). If the mass of the particle is 2 g , the time period of oscillation of the particle is
- A 8 s
- B 6 s
- C 2 s
- D 4 s
Answer & Solution
Correct Answer
(C) 2 s
Step-by-step Solution
Detailed explanation
\(500 F+\pi^2 y=0\) \(\Rightarrow \frac{\mathrm{F}}{\mathrm{m}}+\left(\frac{\pi^2}{500 \mathrm{~m}}\right) \mathrm{y}=0\) \(\therefore \quad a=-\left(\frac{\pi^2}{500 m}\right) y\) \(\qquad ...\mathrm{(i)}\) For SHM, \(a=-\omega^2 y\) \(\qquad ...\mathrm{(ii)}\) Comparing eq(i)…
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