Simple Harmonic Motion

For a body performing simple harmonic motion, the relation between velocity and displacement$v=\omega \sqrt{A^{2}-x^{2}}$Square both side$v^{2}=\omega^{2}(A^{2}-x^{2})$For ( $\omega=1$ )$v^{2}+x^{2}=A^{2}$(a) So, the graph between velocity and displacement will be a circle.(b) We know that,$x=A\sin\;\omega t$Differentiate w.r.t. to $t$$\frac{dx}{dt}=v=+\omega A\cos\omega t$Again differentiate w.r.t. to $t$,$\frac{d^{2}x}{dt^{2}}=a=-\omega^{2}A\sin\;\omega t$From Eq. (i),$a=-\omega^{2}x$Comparing from Eq. $y=mx+c$It is a equation of straight line with slope, $m=-\omega^{2}$.So, the graph between acceleration and displacement will be a straight line.(c) From Eq. (iii),$\frac{d^{2}x}{dt^{2}}=a=-\omega^{2}A\sin\;\omega t$The graph between acceleration and time will be a function of sine wave.(d) From Eq. $(A)$,$v=\omega\sqrt{A^{2}-\omega^{2}}$$v^{2}=\omega^{2}(A^{2}-x^{2})$$x^{2}= \left(-\frac{v^{2}}{\omega^{2}}+A^{2}\right)$$x=\sqrt{A^{2}-\frac{v^{2}}{\omega^{2}}}$ Putting this value in Eq. (iv),$a=-\omega^{2}\sqrt{A^{2}-\frac{v^{2}}{\omega^{2}}}$Square both side$a^{2}=\omega^{4}\left(A^{2}-\frac{v^{2}}{\omega^{2}}\right)$$a^{2}=\omega^{4}A^{2}-v^{2}\omega^{2}$$\left(\frac{a}{\omega^{2}}\right)^{2}=A^{2}-\frac{v^{2}}{\omega^{2}}$$\left(\frac{a}{\omega^{2}}\right)^{2}+\left(\frac{v}{\omega}\right)^{2}=A^{2}$When $\omega \neq 1$, the graph between acceleration $a$ as a function of velocity $v$ in SHM is an ellipse.