The motion of a mass on a spring, with spring constant $\mathrm{K}$ is as shown in figure.
The equation of motion is given by $x(t)=A \sin \omega t+$
$B \cos \omega t$ with $\omega=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$
Suppose that at time $\mathrm{t}=0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t)=\operatorname{Cos}(\omega t-\phi)$, where $C$ and $\phi$ are :
Correct Option: , 4
$x=A \sin \omega t+B \cos \omega t$
$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}-\mathrm{B} \omega \sin \omega \mathrm{t}$
$A t \mathrm{t}=0, \mathrm{x}(0)=\mathrm{B}$
$v(0)=A \omega$
$x=A \sin \omega t+B \sin \left(\omega t+90^{\circ}\right)$
$\mathrm{A}_{\mathrm{net}}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$
$\tan \alpha=\frac{B}{A} \Rightarrow \cot \alpha=\frac{A}{B}$
$\Rightarrow x=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}} \sin (\omega \mathrm{t}+\alpha)$
$\Rightarrow \mathrm{x}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}} \cos (\omega \mathrm{t}-(90-\alpha))$
$x=C \cos (\omega t-\phi)$
$\Rightarrow C=\sqrt{A^{2}+B^{2}}$
$C=\sqrt{\frac{[\mathrm{V}(0)]^{2}}{\omega^{2}}+[x(0)]^{2}}$
$\phi=90-\alpha$
$\tan \alpha=\cos \alpha=\frac{A}{B}$
$\Rightarrow \tan \phi=\frac{\mathrm{v}(0)}{\mathrm{x}(0) \cdot \omega}$
$\phi=\tan ^{-1}\left(\frac{\mathrm{v}(0)}{\mathrm{x}(0) \omega}\right)$
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