Physics - SHM Question with Solution | TestHub

Given here: $\overrightarrow{F}=-k\left(x\hat{i}+y\hat{j}\right) \mathrm{kg} \mathrm{m} {\mathrm{s}}^{-2}$ and $m=1 \mathrm{kg}$

In x-direction, $F_x=-x=m{a}_x$

So, acceleration, $a_x=\frac{d^{2}x}{d{t}^2}=-x$

Now, for particle executing SHM, displacement along x-direction is 

$\Rightarrow x=A_x\sin \left(\omega t+{\varphi }_x\right)$, here, angular frequency, $\omega =1 \mathrm{rad} {\mathrm{s}}^{-1}$ and velocity, $v_x=A_x\omega \cos \left(\omega t+{\varphi }_x\right)$

Given at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{m}$ and $v_x=-\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1}$

So, putting the values, we get

$\frac{1}{\sqrt{2}}=A_x\sin {\varphi }_x$ and $-\sqrt{2}=A_x\cos {\varphi }_x$

From above two equations,

 $\Rightarrow \tan {\varphi }_x=-\frac{1}{2} ...\left(1\right)$

And $A_x=\sqrt{\frac{5}{2}} \mathrm{m} ...\left(2\right)$

Similarly, along y-direction,

$F_y=-y=m{a}_y$,

$\Rightarrow a_y=\frac{d^{2}y}{d{t}^2}=-y$

So, displacement, $y=A_y\sin \left(\omega \mathrm{t}+{\varphi }_{\mathrm{y}}\right)$  and velocity $v_y=A_y\omega \cos \left(\omega t+{\varphi }_y\right)$

Given at $t=0,y=\sqrt{2} \mathrm{m}$ and $v_y=\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1}$

So, putting the values, we get

$\sqrt{2}=A_y\sin {\varphi }_y$ and $\sqrt{2}=A_y\cos {\varphi }_y$

From above two relations, we have

 $\Rightarrow {\varphi }_y=\frac{\pi }{4} ...\left(3\right)$ and $A_y=2 \mathrm{m} ...\left(4\right)$

Now, the value of $\left(x{v}_y-y{v}_x\right)=\sqrt{\frac{5}{2}}\sin \left(\omega t+{\varphi }_x\right)\times 2\cos \left(\omega t+{\varphi }_y\right)-2\sin \left(\omega t+{\varphi }_y\right)\times \sqrt{\frac{5}{2}}\cos \left(\omega t+{\varphi }_x\right)$

$=\sqrt{\frac{5}{2}}\times 2\left(\sin \left(\omega t+{\varphi }_x\right)\cos \left(\omega t+{\varphi }_y\right)-\sin \left(\omega t+{\varphi }_y\right)\times \cos \left(\omega t+{\varphi }_x\right)\right.$

$=\sqrt{10}\sin \left({\varphi }_x-{\varphi }_y\right)$

$=\sqrt{10}\left(\sin {\varphi }_x\cos {\varphi }_y-\cos {\varphi }_x\sin {\varphi }_y\right)$

$=\sqrt{10}\left(\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}\right)\times \frac{1}{\sqrt{2}}\right)$

$=3$