The displacement of a particle executing SHM is given by x = 10 sin [ ωt + (π/3) ]m. The time period of motion is 3.14 s. The velocity of the particle at t = 0 is ______ m/s.

The correct answer is 10 m/s.

We are given the displacement equation of a particle executing Simple Harmonic Motion (SHM):

x=10sin\left(\omega t+\frac{\pi }{3}\right)\text{m}

And the time period: T = 3.14 s

Step 1: Find the angular frequency (ω)

Notice that T = 3.14 s ≈ π s. Using the relation between time period and angular frequency:

\omega =\frac{2\pi }{T}=\frac{2\pi }{\pi }=2\text{rad/s}

Step 2: Find the velocity equation

Velocity is the time derivative of displacement:

v=\frac{dx}{dt}=10\omega cos\left(\omega t+\frac{\pi }{3}\right)

Substituting ω = 2 rad/s:

v=10\times 2\times cos\left(\omega t+\frac{\pi }{3}\right)=20cos\left(\omega t+\frac{\pi }{3}\right)

Step 3: Calculate velocity at t = 0

Substituting t = 0 into the velocity equation:

v|{}_{t=0}=20cos\left(\frac{\pi }{3}\right)

We know that cos\left(\frac{\pi }{3}\right)=cos(60°)=\frac{1}{2}=0.5

Therefore:

v=20\times 0.5=10\text{m/s}

Conclusion: The velocity of the particle at t = 0 is 10 m/s.