Question

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

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

Answer 1

The correct option is A $\sqrt{\frac{3}{2}}$ and $\frac{\pi }{12}$

First SHM is given as $x_1=a\sin wt+a\cos wt=a[\sin wt+\cos wt]$

$\therefore$ $x_1=\sqrt{2}a[\frac{1}{\sqrt{2}}\sin wt+\frac{1}{\sqrt{2}}\cos wt]$

OR $x_1=\sqrt{2}a\left[\cos \right(\pi /4)\sin wt+\sin \pi /4\cos wt]$

$⟹$ $x_1=\sqrt{2}a\sin (wt+\pi /4)$

Second SHM is given as $x_2=a\sin wt+\frac{a}{\sqrt{3}}\cos wt=a[\sin wt+\frac{1}{\sqrt{3}}\cos wt]$

$\therefore$ $x_2=\frac{2}{\sqrt{3}}a[\frac{\sqrt{3}}{2}\sin wt+\frac{\sqrt{3}}{2\sqrt{3}}\cos wt]$

OR $x_2=\frac{2}{\sqrt{3}}a[\frac{\sqrt{3}}{2}\sin wt+\frac{1}{2}\cos wt]$

OR $x_2=\frac{2}{\sqrt{3}}a\left[\cos \right(\pi /6)\sin wt+\sin \pi /6\cos wt]$

$⟹$ $x_2=\frac{2}{\sqrt{3}}a\sin (wt+\pi /6)$

Thus ratio of magnitude $=\frac{\sqrt{2}a}{2a/\sqrt{3}}=\sqrt{\frac{3}{2}}$

Also phase difference $\Delta \varphi =\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$