Question

Find the kinetic energy K of the block as a function of time.

Express your answer in terms of some or all of the variables: k, m, A, and t
To derive the formulas for the major characteristics of motion as functions of time for a horizontal spring oscillator and to practice using the obtained formulas by answering some basic questions.
A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. (Figure 1). Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a system, the equation of motion is

Answer 1

The kinetic energy of the block in simple harmonic motion can be determined using the equation K = 1/2mv^2. In this case, the velocity of the block at any given time can be expressed in terms of the displacement from equilibrium position, x, and the angular frequency, w.

Step-by-step explanation:

The kinetic energy of the block in simple harmonic motion can be determined using the equation K = 1/2mv^2. In this case, the velocity of the block at any given time can be expressed in terms of the displacement from equilibrium position, x, and the angular frequency, w. Using the equation v = Awcos(wt), where A is the amplitude and t is the time, we can substitute v into the kinetic energy equation to obtain K = 1/2mw^2A^2cos^2(wt).

Answer 2

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.