Question

Figure shows snapshots of a particle moving between $+A$ and $-A$ about origin (at $x=0$ ) at different instants. The particle moves in a way that velocity is maximum at $x=0$ and minimum at $x=\pm A$. The correct displacement equation for the motion of the particle is

• A. $A \sin \omega t$
• B. $A \cos \omega t$
• C. $A \tan \omega t$
• D. $A \cot \omega t$

Answer 1

Answer: (B)

As the particle is moving between $+A$ and $-A$ with varying speed about origin (at $x=0$ ) and by observing snapshots we can draw position-time graph for the given motion.
Graph shows a sinusoidal function $x$ with respect to time $t$. From figure, at $t=0$ particle is at $x=+A$ and crosses mean position at $t=T / 4$ and reaches other end in negative direction $(-A)$ at $t=T / 2$.
So, $x(t)=A \cos \omega t$
where, $\omega$ is the angular frequency $=\frac{2 \pi}{T}$.