JEE Main 2021 (22 Jul, Shift 2) Physics Q9: SHM Equations

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 :

Accepted Answer

Option 2: $C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}\ ,\ \phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$

Answer

$C = \sqrt{\frac{2v(0)^2}{\omega^2} + x(0)^2}\ ,\ \phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$

Answer

$C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}\ ,\ \phi = 	an^{-1}\left(\frac{v(0)}{x(0)\omega}ight)$

Answer

$C = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}\ ,\ \phi = 	an^{-1}\left(\frac{x(0)\omega}{v(0)}ight)$

Answer

$C = \sqrt{\frac{2v(0)^2}{\omega^2} + x(0)^2}\ ,\ \phi = 	an^{-1}\left(\frac{x(0)\omega}{2v(0)}ight)$

JEE Main 22 July 2021, Shift 2, Physics Q9

Same idea in other shifts