​ The position of particle oscillation on x-axis is given as x = 10 sin(ωt + π/3).) ​

Question

The position of particle oscillation on x-axis is given as x = 10 sin(\(\omega\)t + \(\frac{\pi}{3}\)). If time period of oscillation is 3.14 second, then displacement of particle at t = 0 is given as \(n\sqrt{3}\) metre , then n is ______

Answer 1

Correct Answer is (5) 

= \(10\times\frac{\sqrt3}{2}= 5\sqrt{3}\)