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 $x\left(t\right)=A\sin \omega t+B\cos \omega t$ with $\omega =\sqrt{\frac{K}{m}}$.

Suppose that at time $t=0,$ the position of mass is $x\left(0\right)$ and velocity $v\left(0\right),$ then its displacement can also be represented as $x\left(t\right)=C\cos (\omega t-\varphi ),$ where $C$ and $\varphi$ are :

• A. $C=\sqrt{\frac{v(0{)}^2}{{\omega }^2}+x(0{)}^2,}\varphi ={\tan }^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$
• B. $C=\sqrt{\frac{2v(0{)}^2}{{\omega }^2}+x(0{)}^2},\varphi ={\tan }^{-1}\left(\frac{x\left(0\right)\omega }{2v\left(0\right)}\right)$
• C. $C=\sqrt{\frac{2v(0{)}^2}{{\omega }^2}+x(0{)}^2,}\varphi ={\tan }^{-1}\left(\frac{x\left(0\right)\omega }{v\left(0\right)}\right)$
• D. $C=\sqrt{\frac{2v(0{)}^2}{{\omega }^2}+x(0{)}^2,}\varphi ={\tan }^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$

Answer 1

The correct option is A $C=\sqrt{\frac{v(0{)}^2}{{\omega }^2}+x(0{)}^2,}\varphi ={\tan }^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$

Given,

$x\left(t\right)=A\sin \omega t+B\cos \omega t$

$\Rightarrow v=\frac{dx}{dt}=A\omega \cos \omega t-B\omega \sin \omega t$

At $t=0,x\left(0\right)=B$ and $v\left(0\right)=A\omega$

$x=A\sin \omega t+B\sin (\omega t+{90}^{\circ })$

Now,


Net amplitude, $A_{net}=\sqrt{A^2+B^2}$

From figure, $\tan \alpha =\frac{B}{A}\Rightarrow \cot \alpha =\frac{A}{B}$

Now, the equation can be written as

$\Rightarrow x=\sqrt{A^2+B^2}\sin (\omega t+\alpha )$

$\Rightarrow x=\sqrt{A^2+B^2}\cos (\omega t-(90-\alpha \left)\right)$

Given, $x=C\cos (\omega t-\varphi )$

$\Rightarrow C=\sqrt{A^2+B^2}$

Substituting the values, $A$ and $B$

$C=\sqrt{\frac{{\left[v\right(0\left)\right]}^2}{{\omega }^2}+{\left[x\right(0\left)\right]}^2}$

$\Rightarrow \varphi ={90}^{\circ }-\alpha$

$\tan \varphi =\cot \alpha =\frac{A}{B}$

$\Rightarrow \tan \varphi =\frac{v\left(0\right)}{x\left(0\right)\omega }$

$\therefore \varphi ={\tan }^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$

Hence, $\left(D\right)$ is the correct answer.