- Home
- Standard 11
- Physics
The motion of a mass on a spring, with spring constant ${K}$ is as shown in figure. The equation of motion is given by $x(t)= A sin \omega t+ Bcos\omega t$ 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
${C}=\sqrt{\frac{2 {v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{x}(0) \omega}{2 {v}(0)}\right)$
${C}=\sqrt{\frac{{v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{x}(0) \omega}{{v}(0)}\right)$
$C=\sqrt{\frac{2 v(0)^{2}}{\omega^{2}}+x(0)^{2}}, \phi=\tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)$
${C}=\sqrt{\frac{{v}(0)^{2}}{\omega^{2}}+{x}(0)^{2}}, \phi=\tan ^{-1}\left(\frac{{v}(0)}{{x}(0) \omega}\right)$
Solution
$x={A} \sin \omega {t}+{B} \sin \omega {t}$
${v} ={A} \sin \omega {t}+{Bcos} \omega {t}$
${{dt}}={A} \omega \cos \omega {t}-{B} \omega \sin \omega {t}$
${At} {t}=0, {x}(0)={B}$
${v}(0)={A} \omega$
${x}={A} \sin \omega {t}+{B} \sin \left(\omega {t}+90^{\circ}\right)$
$A_{\text {net }}=\sqrt{A^{2}+B^{2}}$
$\tan \alpha=\frac{B}{A} \Rightarrow \cot \alpha=\frac{A}{B}$
$\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \sin (\omega t+\alpha)$
$\Rightarrow \quad x=\sqrt{A^{2}+B^{2}} \cos (\omega t-(90-\alpha))$
$x=C \cos (\omega t-\phi)$
$\Rightarrow C=\sqrt{A^{2}+B^{2}}$
$C= \sqrt{\frac{[v(0)]^{2}}{\omega^{2}}+[x(0)]^{2}}$
$\phi= 90-\alpha$
$\tan \alpha=\cos \alpha=\frac{A}{B}$
$\Rightarrow \tan \phi=\frac{v(0)}{x(0) \cdot \omega}$
$\phi= \tan ^{-1}\left(\frac{v(0)}{x(0) \omega}\right)$
