A particle is performing a linear simple harmonic motion if the acceleration and the corresponding velocity of the particle are `a` and `v` respectively. Which of the following graph is correct?

Options

A.
B.
C.
D.

Answer

C

Explanation

If a particle `SHM` with angular frequency of and amplitude `A` then its displacement from mean position will be equal to `x = 4 sin omega t` If its initial phase in equal to zero , velocity will be
`v = (dx)/(dt) = A omega cos omega t`….(i)
and acceleration will be
`a = (dv)/(dt) = -A omega^(2) sin (omega t)`...(ii)
From these equations,
`cos omega t =(v)/(A omega),sin omega t = (a)/(-A omega^(2))`
Squaring and adding
`(v^(2))/(A^(2) omega^(2)) + (a^(2))/(A^(2) omega^(4)) = 1`....(iii)
If `v` taken an `y` axis on x axis then it will be an ellipse.