Question

Two simple harmonic motions are given by:
$x_1=a\sin \omega t+a\cos \omega t$
$x_2=a\sin \omega t+\frac{a}{\sqrt{3}}\cos \omega t$
The ratio of the amplitudes of first and second motion and the phase difference between them are respectively:

• A. $\sqrt{\frac{3}{2}}$ and $\frac{\pi }{6}$
• B. $\frac{\sqrt{3}}{2}$ and $\frac{\pi }{12}$
• C. $\sqrt{\frac{3}{2}}$ and $\frac{\pi }{12}$
• D. $\frac{2}{\sqrt{3}}$ and $\frac{\pi }{12}$

Answer 1

$\Rightarrow x_1=a\sin \omega t+a\cos \omega t$ $=$ $a\sqrt{2}\sin (\omega t+45)$

$\Rightarrow x_2=a\sin \omega t+\frac{a}{\sqrt{3}}\cos \omega t=\frac{2a}{\sqrt{3}}\sin (\omega t+30)$

Using these equations above phasor diagrams are drawn.

Ratio of amplitude $=\frac{a_1}{a_2}=\frac{\sqrt{3}}{\sqrt{2}}$

Phase difference $=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$