## Sunday, May 9, 2021

### Energy of Oscillation: UNIZOR.COM - Physics4Teens - Waves - Mechanical O...

Notes to a video lecture on http://www.unizor.com

Energy of Oscillation

Let's consider an object of mass m on an ideal spring of elasticity k in ideal conditions (no gravity, no friction, no air resistance etc.)

What happens from the energy viewpoint, when we stretch this spring by a distance a from its neutral position?
Obviously, we supply it with some potential energy.

The object on a spring's free end will have this potential energy and, when we let a spring go, a spring will pull the object towards a neutral position, increasing its speed and, therefore, its kinetic energy.

The potential energy, meanwhile, is diminishing since the spring retracts towards its neutral position.
At the moment of crossing the neutral position an object has no potential energy, all its energy is converted into kinetic energy.

When an object moves further, squeezing a spring, it slows down, while squeezing a spring further and further, loses it kinetic energy, but increases potential energy, since a spring is squeezed more and more.

At the extreme position of a squeezed spring all the energy is again potential. An object momentarily stops at this point, having no speed and, therefore, no kinetic energy.

Then the oscillation continues in the opposite direction with similar transformation of energy from potential to kinetic and then back to potential.

Of course, total amount of energy, potential plus kinetic, should remain constant because of the Law of Energy Conservation.

Let's calculate the potential energy we give to an object on a spring by initially stretching a spring by a distance a from its neutral position.

According to the Hooke's Law, stretching a spring by an infinitesimal distance from position x to position x+dx requires a force F(x) proportional to x with a coefficient of proportionality k that depends on the properties of a spring called elasticity.

On the distance dx this force does some infinitesimal amount of work that is equal to
dW(x) = F(x)·dx = k·x·dx.
Integrating this infinitesimal amount of work on a segment from x=0 to x=a, we will obtain the total amount of work W(a) we have to spend to stretch a spring by a distance a from its neutral position.
This amount of work is the amount of potential energy U(a) we supply to an object on a stretched spring.
U(a) = [0,a]k·x·dx = k·a²/2

This formula for potential energy is true for any displacement a. This displacement can be positive (stretching) or negative (squeezing), the potential energy is always positive or zero (for a=0 at the neutral point).

Now we can easily find a speed of an object v0 when it crosses the neutral position. This is the object's maximum speed, since potential energy at this point is zero and all energy is kinetic, which is proportional to a square of its velocity. Its kinetic energy at this point must be equal to the above value U(a). At the same time, if its speed is v0, its kinetic energy is E0=m·v0²/2.
Therefore,
U(a) = m·v0²/2
k·a²/2 = m·v0²/2
|v0| = √k/m·a = ω·a
where ω = √k/m is the same parameter used in expressing the harmonic oscillations in a form x(t)=a·cos(ωt).

In the above expression we used absolute value |v0| because this speed is positive when an object moves from a squeezed position to a stretched one and negative in an opposite direction.

Using this approach we can find an object's velocity vd at any distance d from the neutral position.
The potential energy of an object in this case is U(d)=k·d²/2.
Its kinetic energy is E(d)=m·vd²/2.
Since the total energy is supposed to be equal the potential energy at initial position U(a)=k·a²/2, the kinetic energy equals to
E(d) = U(a) − U(d) =
= k·a²/2 − k·d²/2

From this we can find vd:
m·vd²/2 = k·a²/2 − k·d²/2
|vd| = √(k/m)·(a²−d²)
|vd| = ω·√a²−d²

The above formula for |vd| corresponds to speed being equal to zero at the extreme position of an object at distance a from the neutral point and it being maximum at a neutral point, where d=0.

#### 1 comment:

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