The displacement of a particle executing SHM is given by $x=10 \sin \left(w t+\frac{\pi}{3}\right) m$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $t=0$ is _______ $\mathrm{m} / \mathrm{s}$.

<p>The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:</p> <p>$x = A \sin(\omega t + \phi)$</p> <p>Where:</p> <p>$A$ is the amplitude of the SHM,</p> <p>$\omega$ is the angular frequency,</p> <p>$t$ is the time,</p> <p>$\phi$ is the phase constant (phase angle at $t = 0$).</p> <p>In the given equation, $x = 10 \sin(\omega t + \frac{\pi}{3})$ m, the amplitude $A = 10$ m and the phase constant $\phi = \frac{\pi}{3}$. The time period $T = 3.14$ s is given, from which we can find the angular frequency $\omega$ using the relationship:</p> <p>$\omega = \frac{2\pi}{T}$</p> <p>Substituting the given $T = 3.14$ s:</p> <p>$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$</p> <p>To find the velocity of the particle, we differentiate the displacement $x$ with respect to time $t$. The derivative of the displacement gives the velocity:</p> <p>$v = \frac{dx}{dt}$</p> <p>So, for $x = 10 \sin(\omega t + \frac{\pi}{3})$:</p> <p>$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$</p> <p>Applying differentiation, we get:</p> <p>$v = 10\omega \cos(\omega t + \frac{\pi}{3})$</p> <p>Plug in the value of $\omega = 2$ rad/s and evaluate it at $t = 0$ to find the initial velocity:</p> <p>$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$</p> <p>$v = 20 \cos(\frac{\pi}{3})$</p> <p>$\cos(\frac{\pi}{3}) = \frac{1}{2}$, therefore:</p> <p>$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$</p> <p>Thus, the velocity of the particle at $t = 0$ is $10$ m/s.</p>