Question

For the x-t equation of a particle in SHM along x-axis, answer the following questions using the columns. $x=2+2cos\left(\omega t\right)$ [Given $\overrightarrow{v}$ $\Rightarrow$ velocity, $\overrightarrow{a}$ $\Rightarrow$ acceleration ]
$\begin{array}{cccccc}& ColumnI& & Column2& & Column3\\ \left(I\right)& x=0& \left(i\right)& |\overrightarrow{v}|=0& \left(P\right)& |\overrightarrow{a}|={\omega }^2\\ \left(II\right)& x=2& \left(ii\right)& |\overrightarrow{v}|=\sqrt{3}\omega & \left(Q\right)& |\overrightarrow{a}|=2{\omega }^2\\ \left(III\right)& x=3& \left(iii\right)& |\overrightarrow{v}|=2\omega & \left(R\right)& |\overrightarrow{a}|=zero\\ \left(IV\right)& x=4& \left(iv\right)& |\overrightarrow{v}|=\omega & \left(S\right)& |\overrightarrow{a}|=\sqrt{3}{\omega }^2\end{array}$
​​​​​​​Which of the options is correct for extreme position of this SHM?

• A. (II) (i) (S)
• B. (IV) (iii) (Q)
• C. (I) (i) (Q)
• D. (IV) (i) (R)

Answer 1

The correct option is C (I) (i) (Q)

$x=2+2cos\omega t$
$v=-2\omega sin\omega t$ $\{\therefore v=\frac{dx}{dt}\}$
$a=-2{\omega }^2(x-2)$ $\{\therefore a=\frac{dv}{dt}\}$
$a=-{\omega }^2(x-2)$
$\begin{array}{ccc}mean& extreme& att=\frac{T}{6}\\ x=2& x=0,4& x=3\\ |v|=2\omega & |v|=0& |v|=\sqrt{3}\omega \\ |a|=zero& |a|=2{\omega }^2& |a|={\omega }^2\end{array}$