Question

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

• A. Motion of particle will be on an 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)

Motion will be a simple harmonic motion with amplitude 5.

Given:
$x=3\sin 100\pi t$ and $y=4\sin 100\pi t$

From both equation,

$\frac{x}{3}=\sin 100\pi t...........\left(1\right)$

$\frac{y}{4}=\sin 100\pi t...........\left(2\right)$

from eqs. (1) and (2)

$\Rightarrow \frac{x}{3}=\frac{y}{4}$

$\Rightarrow y=\frac{4}{3}x$

So, this is the equation of a straight line having slope $\frac{4}{3}$.

The equation of resulting motion is

$\overrightarrow{r}=x\hat{i}+y\hat{j}$

$\overrightarrow{r}=(3\hat{i}+4\hat{j})\sin 100\pi t$

So, the amplitude $\left(A\right)$ of the motion,

$A=\sqrt{3^2+4^2}=5$.

Hence, options $\left(b\right)$ and $\left(c\right)$ are the correct answers.