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NEET 2024 Physics Simple Harmonic Motion MCQ Question

Type: MCQ-numerical-Medium-Class 11

If $x = 5 \sin \left( \pi t + \frac{\pi}{3} \right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are:

A

5 cm, 1 s

B

5 m, 1 s

C

5 cm, 2 s

D

5 m, 2 s

Correct Answer

Option D

Detailed Explanation

The amplitude is the coefficient of the sine function, which is 5 m. The angular frequency $\omega = \pi$ , so the time period $T = \frac{2\pi}{\omega} = 2$ s.

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