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Question

Physics Question on Acceleration

If x,vx, v and aa denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period TT, then, which of the following does not change with time?

A

a2T2+4π2v2a^2 T^2 +4 \pi^2 v^2

B

aTx\frac{a T}{x}

C

aT+2πvaT + 2 \pi v

D

aTv\frac{a T}{v}

Answer

aTx\frac{a T}{x}

Explanation

Solution

For a particle executing simple harmonic motion, acceleration, a=ω2xa = - \omega^2 x where ω=2πT\omega = \frac{2 \pi}{T} a=4π2T2xa = - \frac{4 \pi^2}{T^2} x or aTx=4π2T\frac{aT}{x} = - \frac{4 \pi^2}{T} The period of oscillation T is a constant. aTx\therefore \:\: \frac{aT}{x} is a constant.