Question

The resultant amplitude due to superposition of three simple harmonic motions $x_1=3\sin \omega t$, $x_2=5\sin (\omega t+{37}^{\circ })$ and $x_3=-15\cos \omega t$ is:

• A. 18
• B. 10
• C. 12
• D. None of these

Answer 1

The correct option is D None of these

According to the equations given, the amplitude of the motion can be shown as figure (a).

Now resolve the componants of $5$ in x and y axis.

Thus the reultant amplitude $\overrightarrow{A}=(3+5cos{37}^o)\hat{x}-(15-5sin{37}^o)\hat{y}$

$\overrightarrow{A}=(3+5\times 0.8)\hat{x}-(15-5\times 0.6)\hat{y}$

$\overrightarrow{A}=7\hat{x}-12\hat{y}$

$⟹|\overrightarrow{A}|=\sqrt{7^2+{12}^2}=13.89$