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A particle of mass 2 g executes SHM with a period of 12 s and amplitude 10 cm. Find the acceleration of the particle and the restoring force on the particle when it is 2 cm from its mean position. Also find the maximum velocity of the particle. |
Answer» Data : m = 2g = 2 × 10-3 kg, T = 12 s, A = 10 cm = 0.1 m, x = ±2 cm = ± 2 × 10-2 m \(\omega\) = \(\frac{2\pi}T\) = \(\frac{2\times3.142}{12}\) = \(\frac{3.142}{6}\) = 0.5237 rad/s The acceleration of the particle, a = ω2 = (0.52372) (± 2 × 10-2) = ± 0.2743 × 2 × 10-2 = ± 5.486 × 10-3 m/s2 The restoring force on the particle at that position, F = ma = ± (2 × 10-2) (5.486 × 10-3) = ±1.097 × 10-5 N The maximum velocity of the particle, vmax = ωA = 0.5237 × 0.1 5.237 × 10-2 m/s |
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