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: $\phi$ is an acute angle)

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

Answer 1

Answer: (C)

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

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