Question

For the given situation, which of the following options correctly represents the equation for SHM?

• A. $x=A\sin (\omega t+\frac{5\pi }{6})$
• B. $x=A\sin (\omega t+\frac{\pi }{6})$
• C. $x=A\sin (\omega t-\frac{5\pi }{6})$
• D. $x=A\sin (\omega t-\frac{\pi }{6})$

Answer 1

The correct option is A $x=A\sin (\omega t+\frac{5\pi }{6})$

For a particle executing SHM, we can say that,
$x=A\sin (\omega t+\varphi )......\left(1\right)$
From the data given in the figure,
At $t=0,x=\frac{A}{2}\Rightarrow \frac{A}{2}=A\sin \varphi$
$\Rightarrow \varphi =\frac{\pi }{6}\text{or}\frac{5\pi }{6}$

Differentiating $\left(1\right)$ with respect to time on both sides, we get
$v=A\omega \cos (\omega t+\varphi )$
Also, from the diagram, we can deduce that the particle is moving towards the negative extreme position.
Thus, at $t=0$, velocity of the particle is negative.
$\therefore v=A\omega \cos \varphi <0\Rightarrow \cos \varphi <0\Rightarrow \varphi =\frac{5\pi }{6}$
Subitituting $\varphi =\frac{5\pi }{6}$ in $\left(1\right)$, we get
$x=A\sin (\omega t+\frac{5\pi }{6})$
Thus, option (a) is the correct answer.