Question

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

• A. $x=(\omega t+\frac{\pi }{2})$
• B. $x=A\sin (\omega t+\frac{3\pi }{4})$
• C. $x=A\sin (\omega t+\frac{\pi }{4})$
• D. $x=A\sin (\omega t-\frac{3\pi }{4})$

Answer 1

The correct option is B $x=A\sin (\omega t+\frac{3\pi }{4})$


For a particle executing SHM, we can say that,
$x=A\sin (\omega t+\varphi ).......\left(1\right)$
From the data given in the diagram, we can say that at $t=0,x=\frac{A}{\sqrt{2}}$
From $\left(1\right)$, we get
$\frac{A}{\sqrt{2}}=A\sin \varphi \Rightarrow \varphi =\frac{\pi }{4}\text{or}\frac{3\pi }{4}$
On differentiating $\left(1\right)$ with respect to time on both sides, we get
$v=A\omega \cos (\omega t+\varphi )......\left(2\right)$
Also from the diagram, we can see that the particle is moving towards the negative extreme position, so we can say that, at $t=0$, velocity of the particle is negative.
From equation $\left(2\right)$,
$v=A\omega \cos \varphi <0\Rightarrow \cos \varphi <0\Rightarrow \varphi =\frac{3\pi }{4}$
Hence, the equation of the SHM can be written as
$x=A\sin (\omega t+\frac{3\pi }{4})$
Thus, option (b) is the correct answer.