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 $x(t)=C \cos (\omega t-\phi)$, where $C$ and $\phi$ are :

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

Answer 1

Answer: (D)

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

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