Question

The figure shown below represents a particle executing SHM. If the particle starts from the extreme position, $\left(P\right)$, find the correct equation of motion of particle.

• A. $x=5\text{cos}\omega t$
• B. $x=-5\text{cos}\omega t$
• C. $x=5\text{sin}\omega t$
• D. $x=10\text{cos}\omega t$

Answer 1

The correct option is B $x=-5\text{cos}\omega t$

Let ${}^{\prime }{x}^{\prime }$ be the position of particle from mean position
$x=\text{A sin}(\omega t+\varphi )$
Since, the particle starts from point $P$ and amplitude $A=5$
$x=5\text{sin}(\omega t+\varphi )\to \left(1\right)$
$-5=5\text{sin}\left(\omega \right(0)+\varphi )$
$(\because x=-5\text{at}t=0)$
$\Rightarrow \text{sin}\varphi =-1$
$\Rightarrow \varphi =\frac{3\pi }{2}$
$\therefore$ from equation $\left(1\right)$ we get
$x=-5\text{cos}\omega t$
Hence, option (b) is correct.