A body is in simple harmonic motion with time period $$T = 0.5$$ s and amplitude $$A = 1$$ cm. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Assuming the particle starts from the equilibrium position at $$t = 0$$, the time taken to reach $$x = A/2$$ is found using:

$$\frac{A}{2} = A \sin(\omega t)$$

$$\sin(\omega t) = \frac{1}{2}$$

$$\omega t = \frac{\pi}{6}$$

$$t = \frac{\pi}{6 \omega}$$

$$t = \frac{\pi}{6 (2\pi/T)} = \frac{T}{12}$$

$$t = \frac{0.5}{12} \text{ s}$$

The total displacement in this interval is $$\Delta x = x_2 - x_1 = 0.5$$ cm.

$$v_{avg} = \frac{\Delta x}{t}$$

$$v_{avg} = \frac{0.5}{0.5/12}$$

$$v_{avg} = 12 \text{ cm/s}$$