Velocity In Shm

The speed of an SHM oscillator at displacement $x$ is given by $v=\omega \sqrt{A^2-x^2}$, a relation that follows directly from energy conservation between kinetic and potential stores. This expression makes clear that the speed is largest at the centre and shrinks to zero exactly at the turning points where all the energy is potential. Converting units, $A=0.04\text{ m}$, $x=0.02\text{ m}$, and $\omega =50\text{ rad/s}$. Then $v=50\sqrt{(0.04{)}^2-(0.02{)}^2}=50\sqrt{0.0016-0.0004}=50\sqrt{0.0012}=50\times 0.03464\approx 1.73\text{ m/s}$. The value $2.00\text{ m/s}$ is the maximum speed $\omega A$, which would only occur at the mean position where $x=0$. The value $1.00\text{ m/s}$ wrongly uses $\omega x$ instead of the correct square-root expression. The value $0.87\text{ m/s}$ halves the correct answer, a slip from misreading the amplitude. This is the standard NCERT velocity-displacement relation for SHM. As a plausibility check, the speed at $x=A/2$ must be less than the maximum $2.00\text{ m/s}$ but still a large fraction of it, and $1.73\text{ m/s}$, which equals $\frac{\sqrt{3}}{2}$ of the maximum, fits this expectation perfectly.