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A body is vibrating in simple harmonic motion. If its acceleration is 12 cm s−212 cm s-2 at a displacement 3 cm,3 cm, then time period isA6.28 s

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Solution

The formula for acceleration in simple harmonic motion is given by a = -ω²x, where a is acceleration, ω is angular frequency, and x is displacement.

Given that a = 12 cm/s² and x = 3 cm, we can substitute these values into the formula:

12 = -ω² * 3

Solving for ω² gives us ω² = -12 / 3 = -4 s⁻². However, since ω² cannot be negative in real numbers, we take the absolute value, giving us ω² = 4 s⁻².

Taking the square root of both sides gives us ω = 2 rad/s.

The time period T of simple harmonic motion is given by the formula T = 2π / ω.

Substituting ω = 2 rad/s into the formula gives us T = 2π / 2 = π s ≈ 3.14 s.

So, the time period of the body's vibration is approximately 3.14 seconds.

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