Question

In order that the resultant path on superimposing two mutually perpendicular SHM be a circle, the conditions are that:

• A. the amplitudes on both SHM should be equal and they should have a phase difference of $\frac{\pi }{2}$
• B. the amplitudes should be in the ratio $1:2$ and the phase difference should be zero
• C. the amplitudes should be in the ratio $1:2$ and the phase difference should be $\frac{\pi }{2}$
• D. the amplitudes should be equal and the phase difference should be zero

Answer 1

The correct option is A the amplitudes on both SHM should be equal and they should have a phase difference of $\frac{\pi }{2}$

Let the multually perpendicular $\text{SHM}$ be,
$x=A_1\sin \omega t....\left(1\right)$
$\&y=A_2\cos \omega t....\left(2\right)$
Squaring and adding equation $\left(1\right)$ and $\left(2\right)$
$x^2+y^2=(A_1\sin \omega t{)}^2+(A_2\cos \omega t{)}^2$
$\Rightarrow x^2+y^2=A_1^2{\text{sin}}^{2}wt+A_2^2{\text{cos}}^{2}wt$
When $A_1=A_2=A\left(\text{say}\right)$
$\Rightarrow x^2+y^2=A^2({\sin }^2\omega t+{\cos }^2\omega t)$
$\Rightarrow x^2+y^2=A^2$
Comparing the above relation with $x^2+y^2=a^2$, we conclude that it is an equation of circle with radius $=\text{amplitude(A)}$.

Hence resultant path of two mutually perpendicular $\text{SHM}$ on superposition will be circle, if the amplitudes are equal and phase difference between them is $\frac{\pi }{2}$