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The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 ms-1 and 10 ms-2 at a certain instant. Find the amplitude and the time period of the motion.

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 ms-1 and 10 ms-2 at a certain instant. Find the amplitude and the time period of the motion.

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
7 years ago
Sol. Given that, at a particular instant, X = 2cm = 0.02m V = 1 m/sec A = 10 msec–2 We know that a = ω2x ⇒ ω = √(a/×) = √(10/0.02) = √500 = 10√5 T = 2π/ω = 2π/10√5 = 2 × 3.14/10 × 2.236 = 0.28 seconds. Again, amplitude r is given by v = ω (√(r^2-X^2 )) ⇒ v2 = ω2 (r2 – x2) 1 = 500 (r2 – 0.0004) ⇒ r = 0.0489 = 0.049 m ∴ r = 4.9 cm.

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