Question

3 SHMs of equal amplitude A and equal time periods in the same direction combine. The phase of $2^{nd}$ motion is ${60}^{\circ }$ ahead of the first and phase of $3^{rd}$ is ${60}^{\circ }$ ahead of second. Find the amplitude of the resultant.

• A. $A$
• B. $3/2A$
• C. $2A$
• D. $3A$

Answer 1

The correct option is C $2A$

We are given that
$x_1=A\sin \omega t{x}_2=A\sin (\omega t+\frac{\pi }{3})x_3=A\sin (\omega t+\frac{2\pi }{3})$
so the equation of the resultant is
$x_1+x_2+x_3=A(\sin \omega t+\sin (\omega t+\frac{\pi }{3})+\sin (\omega t+\frac{2\pi }{3}))$
adding the first term $\left(\sin \omega t\right)$ and third term $\left(\sin (\omega t+\frac{2\pi }{3})\right)$ using trigonometric formula $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$, we have
$x_1+x_2+x_3=A(2\sin (\omega t+\frac{\pi }{3})\cos \left(\frac{\pi }{3}\right)+\sin (\omega t+\frac{\pi }{3}))\Rightarrow x_1+x_2+x_3=A(2\sin (\omega t+\frac{\pi }{3})\times \frac{1}{2}+\sin (\omega t+\frac{\pi }{3}))\Rightarrow x_1+x_2+x_3=A(\sin (\omega t+\frac{\pi }{3})+\sin (\omega t+\frac{\pi }{3}))\Rightarrow x_1+x_2+x_3=2A\sin (\omega t+\frac{\pi }{3})$
Now we can see that the amplitude of the resultant SHM is $2A$