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

Answer 1

Answer: (A)

The given situation can be shown as

Fon second SHM
Ratio of amplitude $=\frac{a_1}{a_2}=\frac{\sqrt{3}}{\sqrt{2}}$
and phase difference, $\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$