Question

Three S.H.M. of equal amplitude $A$ and equal time period in the same direction combine. The difference in phase between each pair is ${60}^{\circ }$ ahead of the other. The amplitude of the resultant oscillation is

• A. $A$
• B. $2A$
• C. $0$
• D. $4A$

Answer 1

The correct option is B $2A$

Resultant of first two SHM
$R^2=A^2+A^2+2AA\cos {60}^{\circ }=3A^2$
and $\tan {\varphi }_1=\frac{A\sin \varphi }{A+A\cos \varphi }=\frac{A\sin {60}^{\circ }}{A+A\cos {60}^{\circ }}$
$\tan {\varphi }_1=\frac{\frac{\sqrt{3}}{2}A}{A+\frac{A}{2}}=\frac{1}{\sqrt{3}}$
${\varphi }_1={30}^{\circ }$
Phase difference between resultant and 3rd SHM
$={30}^{\circ }+{60}^{\circ }={90}^{\circ }$
$\therefore R^{\prime 2}=3A^2+A^2+\sqrt{3}A.A\cos {90}^{\circ }=4A^2$
$R^{\prime }=2A$