Equation Of Shm

For a body in SHM written as $x=A\sin (\omega t)$, the velocity is the time derivative $v=A\omega \cos (\omega t)$, so its maximum magnitude is $v_{max}=A\omega$, occurring as the body sweeps through the mean position. The velocity leads the displacement by a quarter of a period in phase, reaching its peak exactly when the displacement passes through zero. Reading the given equation, the amplitude is $A=0.05\text{ m}$ and the angular frequency is $\omega =20\pi \text{ rad/s}$. Multiplying gives $v_{max}=0.05\times 20\pi =\pi \text{ m/s}$. The choice $2\pi \text{ m/s}$ wrongly doubles the result, perhaps by misreading the amplitude as $0.1\text{ m}$. The choice $0.05\pi \text{ m/s}$ forgets to include the factor of 20 from the angular frequency. The choice $20\pi \text{ m/s}$ uses only $\omega$ and ignores the amplitude entirely, which is dimensionally the angular frequency, not a speed in this product. This is the standard NCERT result linking maximum velocity to the product of amplitude and angular frequency. As a sanity check, the maximum speed must be finite and proportional to both how far and how fast the reed swings, and $\pi \text{ m/s}\approx 3.14\text{ m/s}$ is physically reasonable for a small rapidly vibrating reed.