Question

Three particles execute SHM along $x-$ direction with same frequency as $x_1=5\sin \omega t$, $x_2=5\sin (\omega t+{53}^{\circ })$, $x_3=-10\cos \omega t$. Find the amplitude of the resultant SHM.

• A. $6\text{units}$
• B. $8\text{units}$
• C. $10\text{units}$
• D. $15\text{units}$

Answer 1

The correct option is C $10\text{units}$

Given,
$x_1=5\sin \omega t.....\left(1\right)$
$x_2=5\sin (\omega t+{53}^{\circ })......\left(2\right)$
$x_3=-10\cos \omega t......\left(3\right)$
We can write $\left(3\right)$ as , $x_3=10\sin (\omega t+{270}^{\circ })$
Finding the resultant amplitude using vector notation.


Resultant amplitude, $|R|=\sqrt{8^2+6^2}=10\text{units}$