A particle is performing a linear simple harmonic motion. If the instantaneous acceleration and velocity of the particle are `a` and `v` respectively, identify the graph which correctly represents the relation between `a` and `v`.

Options

A.
B.
C.
D.

Answer

C

Explanation

If `v = v_(0)sin omega t`, then
`a = (dv)/(dt) = v_(0) omega cos omega t`
`sin omega t = (v)/(v_(0))` …(i)
`cos omega t = (a)/(v_(0)omega)` …(ii)
Squaring and adding these equations, we get
`1 = (v_(2))/(v_(0)^(2) + (a^(2))/(v_(0)omega)`
`:. v^(2) = (-(v_(0))/(omega)) a^(2) + v_(0)^(2)`
Hence, `v^(2)` versus `a^(2)` equation is a straight line with positive intercept and negative slope.