Question

Derive expression for the instantaneous
Kinetic energy potential energy and the total
energy of a Simple harmonic Oscillator:
Show grophically how the potential energy ,
kinetic energy K and the total energy E of a
Simple harmonic oscillator vary with displa-
Cement from equilibrium position.​

Answer 1

Explanation:

Kinetic energy is a simple concept with a simple equation that is simple to derive. Let's do it twice.

Derivation using algebra alone (and assuming acceleration is constant). Start from the work-energy theorem, then add in Newton's second law of motion.

ΔK = W = FΔs = maΔs

Take the the appropriate equation from kinematics and rearrange it a bit.

v2 = v02 + 2aΔs

aΔs = v2 − v02

2

Combine the two expressions.

ΔK = m ⎛

⎝ v2 − v02 ⎞

⎠

2

And now something a bit unusual. Expand.

ΔK = 1 mv2 − 1 mv02

2 2

If kinetic energy is the energy of motion then, naturally, the kinetic energy of an object at rest should be zero. Therefore, we don't need the second term and an object's kinetic energy is just…

K = ½mv2

Derivation using calculus (but now we don't need to assume anything about the acceleration). Again, start from the work-energy theorem and add in Newton's second law of motion (the calculus version).

ΔK = W

ΔK = ⌠

⌡ F(r) · dr

ΔK = ⌠

⌡ ma · dr

ΔK = m ⌠

⌡ dv · dr

dt

Rearrange the differential terms to get the integral and the function into agreement.

ΔK = m ⌠

⌡ dv · dr

dt

ΔK = m ⌠

⌡ dr · dv

dt

ΔK = m ⌠

⌡ v · dv

The integral of which is quite simple to evaluate over the limits initial speed (v) to final speed (v0).

ΔK = 1 mv2 − 1 mv02

2 2

Naturally, the kinetic energy of an object at rest should be zero. Thus an object's kinetic energy is defined mathematically by the following equation…

K = ½mv2