A particle executes simple harmonic motion with an amplitude of $A=5\text{ cm}$ and a time period of $T=2\text{ s}$. What is the maximum velocity of the particle in $\text{cm/s}$?

Solution

1. Identify the given values: amplitude $A=5\text{ cm}$ and time period $T=2\text{ s}$.
2. Calculate the angular frequency using the formula $$\begin{array}{rl}\omega & =\frac{2\pi }{T}\\ & =\frac{2\pi }{2}\\ & =\pi \text{ rad/s}\end{array}$$.
3. Determine the maximum velocity using the formula $v_{\text{max}}=A\omega$.
4. Substitute the values to get $$\begin{array}{rl}{v}_{\text{max}}& =5\times \pi \\ & =5\pi \text{ cm/s}\end{array}$$.
5. Hence the answer is (A).