The motion of a mass on a spring,with spring | vedclass

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(D) Given the equation of motion: $x(t) = A \sin \omega t + B \cos \omega t$.At $t = 0$,$x(0) = A \sin(0) + B \cos(0) = B$. So,$B = x(0)$.The velocity

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

Vedclass · Answer

(D) Given the equation of motion: $x(t) = A \sin \omega t + B \cos \omega t$.At $t = 0$,$x(0) = A \sin(0) + B \cos(0) = B$. So,$B = x(0)$.The velocity is $v(t) = \frac{dx}{dt} = A \omega \cos \omega t - B \omega \sin \omega t$.At $t = 0$,$v(0) = A \omega \cos(0) - B \omega \sin(0) = A \omega$. So,$A = \frac{v(0)}{\omega}$.We want to express $x(t) = A \sin \omega t + B \cos \omega t$ in the form $x(t) = C \cos(\omega t - \phi) = C \cos \omega t \cos \phi + C \sin \omega t \sin \phi$.Comparing the coefficients of $\sin \omega t$ and $\cos \omega t$:$A = C \sin \phi$ and $B = C \cos \phi$.Squaring and adding: $A^2 + B^2 = C^2 (\sin^2 \phi + \cos^2 \phi) = C^2$.Thus,$C = \sqrt{A^2 + B^2} = \sqrt{\left( \frac{v(0)}{\omega} ight)^2 + x(0)^2} = \sqrt{\frac{v(0)^2}{\omega^2} + x(0)^2}$.Dividing the coefficients: $\frac{A}{B} = \frac{C \sin \phi}{C \cos \phi} = 	an \phi$.Therefore,$	an \phi = \frac{A}{B} = \frac{v(0) / \omega}{x(0)} = \frac{v(0)}{x(0) \omega}$,which implies $\phi = 	an^{-1} \left( \frac{v(0)}{x(0) \omega} ight)$.

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