A mass m is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration a corresponding to linear velocity `upsilon` ?

Options

A.
B.
C.
D.

Answer

D

Explanation

In SHM, `y=rsin omegat`
`upsilon=romega cos omegat and a=(dupsilon)/(dt)=-omega^(2)rsinomegat`
`:. cos omegat =(upsilon)/(romega) and sin omegat =-(a)/(omega^(2)r)`
Hence, `cos^(2)omegat + sin^(2)omegat=((upsilon)/(romega))^(2)+[-(a)/(omega^(2)r)]^(2)`
or `1=(upsilon^(2))/(r^(2)omega^(2))+(a^(2))/(omega^(4)r^(2))`
or ` upsilon^(2)=-(a^(2))/(omega^(2))+r^(2)omega^(2)` ...(i0
Comparing (i), with general equation of st. line, `y=mx+c` , we note theat, the equation (i) is a st. line betweent `upsilon^(2) and a^(2)` with negative slope `(1//omega^(2))`. Hence, optiond (d) is correct.