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If a simple harmonic motion is represented by d²x/dt² + αx = 0, its time period is
- 2π/√α
- 2π/α
- 2π√α
- 2πα
Correct answer: 2π/√α
Solution
Comparing d^2x/dt^2 + alpha x = 0 with d^2x/dt^2 + omega^2 x = 0 gives omega = sqrt(alpha). Time period T = 2pi/omega = 2pi/sqrt(alpha).
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