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  • The displacement vector of a particle of mass m is given by r (t)= i A cos omega t + j B sin omega t a) show that the trajectory is an ellipse

The displacement vector of a particle of mass m is given by
r(t)=\widehat{i} A \cos \omega t + \widehat{j}B \sin \omega t

a) show that the trajectory is an ellipse

b) show that F = -m\omega ^{2}r

Answers (1)

For plotting the graph (r-t) or trajectory we relate x & y coordinates.

(a)

x = A \cos \omega t & y = B \sin \omega t

\frac{x }{A}= \cos \omega t          ……… (i)

\frac{y}{B} = \sin \omega t         ……… (ii)

 On squaring and adding (i) & (ii), we get,

\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=\cos ^{2}\omega t +\sin^{2} \omega t

\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1                              

 

Here, we are getting the equation of an ellipse; hence, the trajectory is an ellipse.

(b) Given: x = A \cos \omega t

Therefore, v_{x} = \frac{d_{x}}{dt} = -A\omega \sin \omega t

& a_{x} = \frac{dv_{x}}{dt} = -A\omega^{2} \cos \omega t

Now, y = B \sin \omega t

Therefore v_{y} = \frac{d_{y}}{dt }= B\omega \cos \omega t

& a_{y} = \frac{dv_{y}}{dt }= -B\omega^{2} \sin \omega t

a = a_{x}\widehat{i} + a_{y}\widehat{j}

  = -\widehat{i} A\omega ^{2} \cos\omega t - \widehat{j} B\omega ^{2} \sin\omega t

  = -\omega ^{2} [\widehat{i} A \cos \omega t + \widehat{j} B \sin \omega t]

Therefore, a = - \omega ^{2}\overrightarrow{r} (t)

Thus, the force acting on particle= ma

                                                  = -m \omega ^{2}\overrightarrow{r} (t)

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