Question

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 $\omega =\sqrt{\frac{K}{m}}$ 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 $x(t)=C\cos (\omega t-\varphi )$, where $C$ and $\varphi$ are :

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

Answer 1

Answer: (D)

$x=A\sin \omega t+B\cos \omega t$
$v=\frac{dx}{dt}=A\omega \cos \omega t-B\omega \sin \omega t$
At $t=0,x(0)=B$
$v(0)=A\omega$
$x=A\sin \omega t+B\cos \omega t$

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