Question

The amplitude of the vibrating particle due to superposition of two SHMs: $y_1=\sin (\omega t+\frac{\pi }{3})$ and $y_2=\sin \omega t$ is :

• A. 1
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. 2

Answer 1

The correct option is C $\sqrt{3}$

$y_1=\sin (\omega t+\frac{\pi }{3})$ and $y_2=\sin \omega t$

Resultant is given by $y_1+y_2=\sin (\omega t+\frac{\pi }{3})+\sin \omega t$
$\Rightarrow y_1+y_2=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t+\sin \omega t=\frac{3}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t$
And we know that amplitude of $y=a\sin \omega t+b\cos \omega t$ is $\sqrt{a^2+b^2}$

So the amplitude of the resultant is: $\sqrt{{\left(\frac{3}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{\frac{12}{4}}=\sqrt{3}$