Shm Kinematics

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A particle executing simple harmonic motion has amplitude 5 cm and passes through the mean position, so what is true about its speed and acceleration there?

RankGuru · Accepted Answer

Consider the defining relations of simple harmonic motion given in NCERT Class 11, Chapter 14 (Oscillations): the velocity is $v = \omega\sqrt{A^2 - x^2}$ and the acceleration is $a = -\omega^2 x$, where $x$ is displacement from the mean position. At the mean position $x = 0$, so the velocity becomes $v = \omega\sqrt{A^2 - 0} = \omega A$, which is its greatest value, while the acceleration becomes $a = -\omega^2 (0) = 0$. Physically, at the centre the restoring force vanishes because there is no displacement, so nothing decelerates the particle and it moves fastest. The option claiming zero speed and maximum acceleration describes the extreme positions, not the mean, so it is wrong. The option claiming both are maximum is impossible because velocity peaks where acceleration is zero and vice versa. The option that both are zero would describe a particle permanently at rest, contradicting oscillation. A quick plausibility check: with $A = 5$ cm, $v_{max} = \omega A$ has units of velocity and $a = 0$ has units of acceleration, both consistent, and the energy picture agrees since all energy is kinetic at the centre.

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