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\left(t\right)=A\sin \omega t+$$B\cos \omega \mathrm{t}$ with $\omega =\sqrt{\frac{K}{m}}$.
Suppose that at time $t=0,$the position of mass is $x\left(0\right)$ and velocity $v\left(0\right),$ then its displacement can also be represented as $x\left(t\right)=C\cos (\omega t-\varphi ),$ where $C$ and $\varphi$ are 

• A

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

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

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

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

Hence, the correct answer is option: 'D'.