JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 17)
The motion of a mass on a spring, with spring constant K is as shown in figure.
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The equation of motion is given by
x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$
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 :
The equation of motion is given by
x(t) = A sin$$\omega$$t + B cos$$\omega$$t with $$\omega$$ = $$\sqrt {{K \over m}} $$
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 :
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {2v(0)}}} \right)$$
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{x(0)\omega } \over {v(0)}}} \right)$$
$$C = \sqrt {{{v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
$$C = \sqrt {{{2v{{(0)}^2}} \over {{\omega ^2}}} + x{{(0)}^2}} ,\phi = {\tan ^{ - 1}}\left( {{{v(0)} \over {x(0)\omega }}} \right)$$
Explanation
$$C\cos \phi = x(0)$$
$$tC\omega \sin \phi = v(0)$$
$${\left[ {{{v(0)} \over \omega }} \right]^2} + {[x(0)]^2} = {C^2}$$
$$\tan \phi = {{v(0)} \over {x(0)\omega }}$$
$$tC\omega \sin \phi = v(0)$$
$${\left[ {{{v(0)} \over \omega }} \right]^2} + {[x(0)]^2} = {C^2}$$
$$\tan \phi = {{v(0)} \over {x(0)\omega }}$$
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