A particle is doing simple harmonic motion of amplitude $0.06 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum velocity of the particle is _________ $\mathrm{cm} / \mathrm{s}$.

<p>For a particle performing simple harmonic motion (SHM), the maximum velocity $v_{max}$ can be calculated using the formula: <p>$v_{max} = A\omega$</p> <p>where $A$ is the amplitude of the motion and $\omega$ is the angular frequency. The angular frequency $\omega$ is related to the time period $T$ by the formula:</p> <p>$\omega = \frac{2\pi}{T}$</p> <p>Given:</p> <ul> <li>Amplitude, $A = 0.06 \, \mathrm{m}$</li><br> <li>Time period, $T = 3.14 \, \mathrm{s}$</li> </ul> <p>First, we find the angular frequency:</p> <p>$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14}$</p> <p>Substituting $\omega$ and $A$ in the formula for $v_{max}$:</p> <p>$v_{max} = A\omega = 0.06 \times \frac{2\pi}{3.14}$</p> <p>$v_{max} = 0.06 \times \frac{2 \times 3.14}{3.14}$</p> <p>$v_{max} = 0.06 \times 2$</p> <p>$v_{max} = 0.12 \, \mathrm{m/s}$</p> <p>To convert meters per second to centimeters per second, we use the conversion factor $1 \, \mathrm{m/s} = 100 \, \mathrm{cm/s}$. Therefore,</p> <p>$v_{max} = 0.12 \, \mathrm{m/s} \times 100 \, \mathrm{cm/m} = 12 \, \mathrm{cm/s}$</p> <p>Thus, the maximum velocity of the particle is $12 \, \mathrm{cm/s}$.</p></p>