Question

What will be the displacement equation of the simple harmonic motion obtained by combining the motions?
$x_1=2\sin \omega t,x_2=4\sin \left(\omega t+\frac{\pi }{6}\right)$ and $x_3=6\sin \left(\omega t+\frac{\pi }{3}\right)$
(Note: $\varphi$ is an acute angle)

• A. $x=10.25\sin \left(\omega t+\varphi \right)$
• B. $x=10.25\sin \left(\omega t-\varphi \right)$
• C. $x=11.25\sin \left(\omega t+\varphi \right)$
• D. $x=11.25\sin \left(\omega t-\varphi \right)$

Answer 1

Answer: (C)

The resultant equation is
$x=A\sin \left(\omega t+\varphi \right)$
$\sum A_x=2+4\cos 30^{\circ }+6\cos 60^{\circ }=8.46$

And $\sum A_y=4\sin 30^{\circ }+6\cos 30^{\circ }=7.2$
$\therefore A=\sqrt{{\left(\sum A_x\right)}^2+{\left(\sum A_y\right)}^2}$
$=\sqrt{{\left(8.46\right)}^2+{\left(7.2\right)}^2}=11.25$
And $\tan \varphi =\frac{\sum A_y}{\sum A_x}=\frac{7.2}{8.46}=0.85$
$\Rightarrow \varphi ={\left(\tan \right)}^{-1}\left(0.85\right)=40.4^{\circ }$
Thus, the displacement equation of combined motion is
$x=11.25\sin \left(\omega t+\varphi \right)$
Where, $\varphi =40.4^{\circ }$