Question

A particle is subjected to two SHMs simultaneously $x_1=a_1\sin \omega t$ and $x_2=a_{2}sin(\omega t+\varphi )$ Where $a_1=3.0cm$, $a_2=4.0cm$. Find resultant amplitude if the phase difference $\varphi$ has values(a) 0 (b) 60 (c) 90

Answer 1

$x_1=a_1\sin \omega t$

$x_2=a_2\sin (\omega t+\theta )$

Given that $a_1=30cm$ and $a_2=4.0cm$

$1$) In this case $\theta =0$

$x_1=3\sin \omega t$

$x_2=4\sin (\omega t+\theta )$

So, amplitude of resultant motion simply add p which is $(3+4)=7cm$

$2$) $x_1=3\sin \omega t$

$x_2=4\sin (\omega t+{60}^0)=4\sin \omega t\cos {60}^0+4\sin {60}^0\cos \omega t$

Or, $x_2=2\sin \omega t+2\sqrt{3}\cos \omega t$

Resultant motion should be $x=x_1+x_2=3\sin \omega t+2\sin \omega t+2\sqrt{3}\cos \omega t$

Or, $x=5\sin \omega t+2\sqrt{3}\cos \omega t$

The resultant of two perpendicular S.H.M. So, resultant amplitude is

$=\sqrt{5^2+(2\sqrt{3}{)}^2}=\sqrt{25+12}=\sqrt{37}cm$

$3$) $x_1=3\sin \omega t$

$x_2=4\sin (\omega t+{90}^0)=4\cos \omega t$

So, $x=x_1+x_2$

$x=3\sin { \omega t } +4\cos { \omega t } $

So, resultant amplitude $=\sqrt{3^2+4^2}=5cm$ As the resultant motion is superposition of two perpendicular S.H.M

Answers are :$7cm,\sqrt{37}cm,5cm$