Question

A particle is subjected to two simple harmonic motion along $x$ and $y$ directions, according to equation $x=3\sin 100\pi t;y=4\sin 100\pi t$. Then,

• A. motion of particle will be on ellipse travelling in clockwise direction.
• B. Motion of particle will be on a straight line with slope $\frac{4}{3}$
• C. Motion will be a simple harmonic motion with amplitude $5$.
• D. Phase difference between two motions is $\frac{\pi }{2}$.

Answer 1

Answer: (B), (C)

The correct options are
B Motion of particle will be on a straight line with slope $\frac{4}{3}$
C Motion will be a simple harmonic motion with amplitude $5$.

Given
$x=3\sin 100\pi t$
$y=4\sin 100\pi t$
Equation of path is
$\frac{y}{x}=\frac{4}{3}$
i.e. $y=\frac{4}{3}x$
which is equation of a straight line having slope $\frac{4}{3}$
Equation of resulting motion is $\overrightarrow{r}=x\hat{i}+y\hat{j}$
$\overrightarrow{r}=(3\hat{i}+4\hat{j})\sin 100\pi t$
Amplitude is $\sqrt{3^2+4^2}=5$

Why this question? When particle undergoes multiple non collinear SHMs, a convinient way is to use vector analysis.