1. Given SHM equation

The displacement is $x(t)=A\sin \omega t + B\cos \omega t$ with $\omega=\sqrt{\frac{K}{m}}.$

We are told that at $t=0$, position and velocity are $x(0)$ and $v(0)$ respectively.

2. Use initial conditions to find $A$ and $B$

From $x(t)=A\sin \omega t + B\cos \omega t,$ at $t=0$: $x(0)=A\sin 0 + B\cos 0 = B.$ So, $B=x(0).$

Now differentiate to get velocity: $v(t)=\frac{dx}{dt}=A\omega \cos \omega t - B\omega \sin \omega t.$ At $t=0$: $v(0)=A\omega \cos 0 - B\omega \sin 0 = A\omega.$ Thus, $A=\frac{v(0)}{\omega}.$

So the motion becomes $x(t)=\frac{v(0)}{\omega}\sin \omega t + x(0)\cos \omega t.$

3. Compare with the form $x(t)=C\cos(\omega t-\phi)$

Using the identity $\cos(\omega t-\phi)=\cos \omega t\cos\phi + \sin \omega t\sin\phi,$ we get $x(t)=C\cos(\omega t-\phi)=C\cos\phi\cos\omega t + C\sin\phi\sin\omega t.$

Comparing with $x(t)=x(0)\cos\omega t + \frac{v(0)}{\omega}\sin\omega t,$ we obtain: $C\cos\phi=x(0),$ $C\sin\phi=\frac{v(0)}{\omega}.$

4. Find $C$

Square and add: $C^2\cos^2\phi + C^2\sin^2\phi = x(0)^2 + \frac{v(0)^2}{\omega^2}.$ Since $\sin^2\phi+\cos^2\phi=1$, $C^2=x(0)^2+\frac{v(0)^2}{\omega^2}.$ Hence, $C=\sqrt{x(0)^2+\frac{v(0)^2}{\omega^2}}.$

5. Find $\phi$

Divide the two equations: \tan\phi=\frac{C\sin\phi}{C\cos\phi}=rac{\frac{v(0)}{\omega}}{x(0)}=rac{v(0)}{x(0)\omega}. Therefore, $\phi=\tan^{-1}\left(\frac{v(0)}{x(0)\omega}\right).$

6. Match with options

We found: $C=\sqrt{\frac{v(0)^2}{\omega^2}+x(0)^2}, \qquad \phi=\tan^{-1}\left(\frac{v(0)}{x(0)\omega}\right).$

This matches Option C.

7. Comparison with stored answer

Stored correct answer: C

My derived answer: C

So they agree.