A) The path of the particle is an ellipse
B) The velocity and acceleration of the particle are normal to each other at \[t=\pi /(2p)\]
C) The acceleration of the particle is always directed towards a focus
D) The distance travelled by the particle in time interval \[t=0\] to \[t=\pi /(2p)\] is \[a\]
Correct Answer: A
Solution :
\[x=a\cos (pt)\] and \[y=b\sin (pt)\] (given) \[\therefore \] \[\cos pt=\frac{x}{a}\] and \[\sin pt=\frac{y}{b}\] By squaring and adding \[{{\cos }^{2}}(pt)+{{\sin }^{2}}(pt)=\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Hence path of the particle is ellipse. Now differentiating x and y w.r.t. time \[{{v}_{x}}=\frac{dx}{dt}=\frac{d}{dt}(a\cos (pt))=-ap\sin (pt)\] \[{{v}_{y}}=\frac{dy}{dt}=\frac{d}{dt}(b\sin (pt))=bp\cos (pt)\] \[\therefore \ \ \vec{v}={{v}_{x}}\hat{i}+{{v}_{y}}\hat{j}=-ap\sin (pt)\hat{i}+bp\cos (pt)\hat{j}\] Acceleration \[\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}[-ap\sin (pt)\hat{i}+bp\cos (pt)\hat{j}]\] \[\vec{a}=-a{{p}^{2}}\cos (pt)\ \hat{i}-b{{p}^{2}}\sin (pt)\hat{j}\] Velocity at \[t=\frac{\pi }{2p}\] \[\vec{v}=-ap\sin p\left( \frac{\pi }{2p} \right)\ \hat{i}+bp\cos p\left( \frac{\pi }{2p} \right)\hat{j}\]\[=-ap\ \hat{i}\] Acceleration at \[t=\frac{\pi }{2p}\] \[\vec{a}=a{{p}^{2}}\cos p\left( \frac{\pi }{2p} \right)\ \hat{i}-b{{p}^{2}}\sin p\left( \frac{\pi }{2p} \right)\hat{j}\]\[=-b{{p}^{2}}\hat{j}\] As \[\vec{v}\ .\ \vec{a}=0\] Hence velocity and acceleration are perpendicular to each other at \[t=\frac{\pi }{2p}\].
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