Question 38: Oscillations

Question

A particle of mass $1 \mathrm{~kg}$ is subjected to a force which depends on the position as $\vec{F}=$ $-k(x \hat{\imath}+y \hat{\jmath}) \mathrm{kg}\, \mathrm{m} \mathrm{s}^{-2}$ with $k=1 \mathrm{~kg} \mathrm{~s}^{-2}$. At time $t=0$, the particle's position $\vec{r}=$ $\left(\frac{1}{\sqrt{2}} \hat{\imath}+\sqrt{2} \hat{\jmath}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{\imath}+\sqrt{2} \hat{\jmath}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$. Let $v_{x}$ and $v_{y}$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5 \mathrm{~m}$, the value of $\left(x v_{y}-y v_{x}\right)$ is __________ $m^{2} s^{-1}$.

Accepted Answer

$$
F_x=-x=m a_x
$$
So $a_x=\frac{d^2 x}{d t^2}=-x$
$$
\Rightarrow x=A_x \sin \left(\omega t+\phi_xight) \quad(\omega=1 \mathrm{rad} / \mathrm{s})
$$
and $v_x=A_x \omega \cos \left(\omega t+\phi_xight)$
at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}$ and $v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\frac{1}{\sqrt{2}}=A_x \sin \phi_x$
and $-\sqrt{2}=A_x \cos \phi_x$
$$
\begin{aligned}
& \Rightarrow 	an \phi_x=-\frac{1}{2} ......(1) \
& 	ext { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2)
\end{aligned}
$$
Similarly
$$
\begin{aligned}
& F_y=-y=m a_y . \\
& \Rightarrow \frac{d^2 y}{d t^2}=-y
\end{aligned}
$$
So, $y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}ight) \quad(\omega=1 \mathrm{rad} / \mathrm{s})$ and
$v_y=A_y \omega \cos \left(\omega t+\phi_yight)$
at $t=0,y=\sqrt{2} \mathrm{~m}$ and $v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\sqrt{2}=A_y \sin \phi$
and $\sqrt{2}=A_y \cos \phi$
$$
\Rightarrow \phi=\frac{\pi}{4} 	ext { and } A_y=2
$$
So,
$\left(x v_y-y v_xight) $
$
 =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_xight) 	imes 2 \cos \left(\omega t+\phi_yight)-2 \sin \left(\omega t+\phi_yight) 	imes \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_xight) $
$ =\sqrt{\frac{5}{2}} 	imes 2\left(\sin \left(\omega t+\phi_xight) \cos \left(\omega t+\phi_yight)-\sin \left(\omega t+\phi_yight) 	imes \cos \left(\omega t+\phi_xight)ight. $
$ =\sqrt{10} \sin \left(\phi_x-\phi_yight)$
$ =\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_yight) $
$ =\sqrt{10}\left(\frac{1}{\sqrt{5}} 	imes \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}ight) 	imes \frac{1}{\sqrt{2}}ight) $
$=3$

Oscillations

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