Final answer:
The kinetic energy K of a block in SHM at any time t is K(t) = \frac{1}{2}kA^2 \sin^2(\omega t), with \omega being the angular frequency \sqrt{k/m}. It oscillates between 0 and \frac{1}{2}kA^2.
Step-by-step explanation:
To find the kinetic energy K of the block as a function of time in a system undergoing simple harmonic motion (SHM), we can use the energy transformation principles of SHM. The maximum kinetic energy occurs when the block passes through the equilibrium position (x=0) and is equal to the maximum potential energy stored in the spring when the block is at its maximum displacement (±A).
The kinetic energy K of the block at any time t is given by:
K = \frac{1}{2}mv^2
Since the velocity v in SHM is given by v = -Aw \sin(\omega t) where \omega = \sqrt{k/m}, we can rewrite kinetic energy as:
K(t) = \frac{1}{2}m(A^2\omega^2 \sin^2(\omega t))
K(t) = \frac{1}{2}kA^2 \sin^2(\omega t)
Here, \omega is the angular frequency of the oscillation. Therefore, the kinetic energy as a function of time, K(t), oscillates between 0 and \frac{1}{2}kA^2.