TS EAMCET · Physics · Gravitation

A simple harmonic oscillator has an amplitude of $0.5 m$ and a time period of $2$ seconds. What is the magnitude of acceleration when it is displaced from the mean position by $0.25 m ?$

A $π^{2} m s^{- 2}$
B $\frac{π^{2}}{2} m s^{- 2}$
C $\frac{π^{2}}{4} m s^{- 2}$
D $\frac{π^{2}}{8} m s^{- 2}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{π^{2}}{4} m s^{- 2}$

Step-by-step Solution

Detailed explanation

Magnitude of acceleration in terms of displacement is given by, a=ω2x ⇒a=2πT2x ⇒a=2π22×0.25=π24

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A simple harmonic oscillator has an amplitude of 0.5 m and a time period of 2 seconds. What is the magnitude of acceleration when it is displaced from the mean position by 0.25 m?

Step-by-step Solution

A simple harmonic oscillator has an amplitude of $0.5 m$ and a time period of $2$ seconds. What is the magnitude of acceleration when it is displaced from the mean position by $0.25 m ?$