MCQ
A vibratory motion is represented by $x = 2A\,\cos \omega t + A\,\cos \,\left( {\omega t + \frac{\pi }{2}} \right) + A\,\cos \,\left( {\omega t + \pi } \right)$ $ + \frac{A}{2}\,\cos \left( {\omega t + \frac{{3\pi }}{2}} \right)$. The resultant amplitude of the motion is
- A$\frac{{9A}}{2}$
- ✓$\frac{{\sqrt 5 A}}{2}$
- C$\frac{{5A}}{2}$
- D$2A$