Question 117 of 210

The motion of a mass on a spring, with spring constant $K$ is as shown in figure.

The equation of motion is given by, $x(t)=A\sin \omega t+$$B\cos \omega t$ with $\omega =\sqrt{\dfrac{K}{m}}$.
Suppose that at time $t=0,$the position of mass is $x(0)$ and velocity $v(0),$ then its displacement can also be represented as $x(t)=C\cos (\omega t-\phi ),$ where $C$ and $\phi$ are

A$C=\sqrt{\dfrac{2v(0{)}^{2}}{{\omega }^{2}}+x{(0)}^{2}},\phi ={\tan }^{-1}\left(\dfrac{v(0)}{x(0)\omega }\right)$B$C=\sqrt{\dfrac{2v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\tan }^{-1}\left(\dfrac{x(0)\omega }{2v(0)}\right)$C$C=\sqrt{\dfrac{v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\tan }^{-1}\left(\dfrac{x(0)\omega }{v(0)}\right)$D$C=\sqrt{\dfrac{v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\tan }^{-1}\left(\dfrac{v(0)}{x(0)\omega }\right)$

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Answer: (D) $C=\sqrt{\dfrac{v(0{)}^{2}}{{\omega }^{2}}+x(0{)}^{2}},\phi ={\tan }^{-1}\left(\dfrac{v(0)}{x(0)\omega }\right)$

$x=Asin\omega t+Bcos\omega t$
$v=\dfrac{dx}{dt}=A\omega \cos \omega t-B\omega \sin \omega t$
At $t=0,x(0)=B$
$v(0)=A\omega$
$x=A\sin \left(\omega +B\sin \left(\omega t+90^{\circ}\right)\right.$

${A}_{net}=\sqrt{{A}^{2}+{B}^{2}}$
$\tan \alpha =\dfrac{B}{A}\Rightarrow \cot \alpha =\dfrac{A}{B}$
$\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\sin (\omega t+\alpha )$
$\Rightarrow x=\sqrt{{A}^{2}+{B}^{2}}\cos (\omega t-(90-\alpha ))$
$x=C\cos (\omega t-\phi )$
$\Rightarrow C=\sqrt{{A}^{2}+{B}^{2}}$
$C=\sqrt{\dfrac{[v(0){]}^{2}}{{\omega }^{2}}+[x(0){]}^{2}}$
$\phi =90-\alpha$
$\tan \alpha =\cos \alpha =\dfrac{A}{B}$
$\Rightarrow \tan \phi =\dfrac{v(0)}{x(0)\cdot \omega }$
$\phi ={\tan }^{-1}\left(\dfrac{v(0)}{x(0)\omega }\right)$