Question

A particle is subjected to three SHMs in the same direction simultaneously each having equal amplitude a and equal time period. The phase of the second motion is 30 ahead of the first and the phase of the third motion is 30 ahead of the second. Find the amplitude of the resultant motion.

Answer 1

Here first ahead of first by ${30}^0$ and third motion is ahead of second by ${30}^0$. So, the third motion is ahead of first motion by ${60}^0$. Let amplitude of the motion is $a$ and time period of motion is $T$.

So, angular frequency $\omega =\frac{2\pi }{T}$

Equation for displacement of first motion, second motion and third motion respectively are

$x_1=a\sin \omega t$

$x_2=a\sin (\omega t+{30}^0)$

$x_3=a\sin (\omega t+{30}^0)$

So the resultant motion will be: $x=x_1+x_2+x_3$

$\Rightarrow x=a\sin \omega t+a\sin (\omega t+{30}^0)+a\sin (\omega t+{30}^0)$

$\Rightarrow x=a\sin \omega t+a[\sin \omega t\cos {30}^0+\sin {30}^0\cos \omega t]+a[\sin \omega t\cos {60}^0+\sin {60}^0\cos \omega t]$

$\Rightarrow x=a\sin \omega t+a[\frac{\sqrt{3}}{2}\sin \omega t+\frac{1}{2}\cos \omega t]+a[\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t]$

$\Rightarrow x=a\sin \omega t+a[1+\frac{\sqrt{3}}{2}+\frac{1}{2}]\sin \omega t+a[\frac{\sqrt{3}}{2}+\frac{1}{2}]\cos \omega t$

$\Rightarrow x=\left(\frac{\sqrt{3+3}}{2}\right)a\sin \omega t+\left(\frac{\sqrt{3}+1}{2}\right)a\cos \omega t$

This equation shows that the resultant motion has two perpendicular S.H.M as its component. So, the resultant amplitude is

$A=\sqrt{(\frac{\sqrt{3+3}}{2}{)}^2{a}^2+(\frac{\sqrt{3}+1}{2}{)}^2{a}^2}$

$\Rightarrow A=\frac{a}{2}\sqrt{(\sqrt{3}+3{)}^2+(\sqrt{3}+1{)}^2}$

$\Rightarrow A=\frac{a}{2}\sqrt{3+6\sqrt{3}+9+3+2\sqrt{3}+1}$

$\Rightarrow A=\frac{a}{2}=\sqrt{16+8\sqrt{3}}$

$\Rightarrow A=(4+2\sqrt{3})a$