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Q.

A particle mass ‘m’ free to move in x-y plane is subjected to a force whose components are F_x= –kx and F_y= –ky, where ‘k’ is a constant. The particle is released when t = 0 at the point (2,3). The particle is in simple harmonic motion along the line represented by the equation.

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a

$\large 2x - 3y\, = \,0$

b

$\large 3x - 2y\, = \,0$

c

$\large 3x - y\, = \,0$

d

$\large x - 2y\, = \,0$

answer is B.

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Detailed Solution

Since $\large {a_x}\, \propto \,\left( { - x} \right)$ and $\large {a_y}\, \propto \,\left( { - y} \right)$ The particle will execute SHM parallel to x-axis and y-axis.

Let $\large x\, = \,A\cos \left( {\omega t + {\phi _1}} \right)$

and $\large y\, = \,B\cos \left( {\omega t + {\phi _2}} \right)$

At t= 0, x= 2 and $\large \frac{{dx}}{{dt}}\, = \,0$

$\large \therefore \,\frac{{dx}}{{dt}}\, = \, - A\omega \sin \left( {\omega t + {\phi _1}} \right)\, = \,0$

$\large \Rightarrow \,A\omega \sin {\phi _1}\, = \,0\,\,\,\,\, \Rightarrow \,{\phi _1}\, = \,0$

Also, $\large 2\, = \,A\cos \left( {\omega \times 0 + 0} \right)\,\, \Rightarrow \,A\, = 2$

Similarly at t= 0, y = 3 and $\large \frac{{dy}}{{dt}}\, = \,0$

$\large \therefore \,{\phi _2}\, = \,0\,\,and\,\,B\, = \,3$

$\large \therefore \,x\, = \,2\cos \omega t\,\,and\,\,y\, = \,3\,\cos \omega t$

$\large \therefore \,\frac{y}{x}\, = \,\frac{3}{2}\,\, \Rightarrow \,3x - 2y\, = \,0$

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