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

__Energy of Spring__

The energy discussed in the previous introductory lecture was an energy related to constant force acting on an object. In this lecture we will address a more difficult case of a variable force of a spring acting horizontally on an object according to the Hooke's Law.

Let's remind the Hooke's Law. It states that the force of a spring, fixed on one end, onto a point-object attached to its free end is proportional to the length the spring is extended or compressed from its neutral position.

The corresponding formula for the force

**of a spring onto an object is**

*F*

*F = −k·L*where

**is a displacement of the spring's free end and**

*L***is a**

*k**coefficient of elasticity*- a specific characteristic of "stiffness" of each spring.

The minus sign signifies that the direction of the force is opposite to a displacement. The frame of reference is associated with the neutral position of a spring and the X-axis is directed towards spring's stretching. Then, for positive displacement

**(stretching) the force is directed towards neutral point (that is, negative), while for negative displacement (compression) the force also directed towards a neutral position, (that is, positive).**

*L*At this point we recommend to refresh the material in the "Work and Elasticity" lecture of the "Mechanical Work" chapter of the "Mechanics" part of this course.

As stated in that lecture, the total amount of work

**to compress (or stretch) a spring with elasticity**

*W***by length**

*k***equals**

*L*

*W = k·L²/2*regardless of how exactly we compress or stretch a spring.

Assume now that we have compressed a spring by the length

**, performing work equaled to**

*L***. At a spring's end we have attached an object of mass**

*W=k·L²/2***. Then we let the spring go by itself with the object attached to its end.**

*m*Intuitively, we think that, at first, the spring should return back to its neutral position. During this time the object's speed will increase because the force of elasticity of a spring will push it.

But then, when a spring reaches its neutral position, the object, having certain speed, will continue its move by inertia, stretching the spring. During this time the force of elasticity of a spring will slow down the movement of the object until it stops at the end of the stretching process.

At that time the force of elasticity will pull the object back to a neutral position, increasing its speed.

At the neutral position object continues moving by inertia, compressing a string, until the force of elasticity stops its movement when the spring will be in the compressed position, as in the beginning of our process.

At this point the process repeats itself and can continue like this indefinitely.

Well, our intuition is correct, a spring with an object attached to it will oscillate indefinitely. Let's analyze quantitatively these oscillations.

Our first goal is to find the dependency of the object's position on time. This can be done using the Newton's Second Law, taking into account the initial position of the object at the end of a compressed spring with no initial speed.

Let

**be a function describing the position of the object at the end of a spring relatively to its neutral position with stretching being a positive direction. Then**

*S(t)***and**

*S(0)=−L***. The acceleration is the second derivative of**

*S'(0)=0***, that is**

*S(t)***.**

*a=S"(t)***as**

*S(t)***. Now the Newton's Second Law**

*F=−k·S(t)***looks like this differential equation:**

*F=m·a*

*−k·S(t) = m·S"(d)*with two initial conditions:

*S(0) = −L*

*S'(0) = 0*This differential equation is fully solvable and its general solution is

*x(t) = C*

+ C_{1}·cos(t·√k/m) ++ C

_{2}·sin(t·√k/m)Constants

**and**

*C*_{1}**can be determined using the initial conditions:**

*C*_{2}

*C*_{1}= −L

*C*_{2}= 0Therefore,

*S(t) = −L·cos(t·√k/m)*This function describes simple sinusoidal oscillations with amplitude

**and period**

*L***.**

*T=2π√m/k*We can also calculate the speed of an object as the first derivative of distance:

*V(t) = S'(t) =**d*

**S(t)/**d**t =**

= L·√k/m·sin(t·√k/m)= L·√k/m·sin(t·√k/m)

During the first quarter of this time an object accelerates under the influence of the elasticity of a spring, moving from the extremely compressed position on a spring to its neutral position. During the next quarter of the period the spring's elasticity slows it down, while it moves to extremely stretched position of a spring. During the next quarter of the period elasticity of a spring accelerates it again to a string's neutral position. Finally, the elasticity slows the object down, as it moves to the initial most compressed position.

Let's calculate the work that spring performs during the first quarter of its period from complete compression to a neutral point by accelerating the object from initial speed

**to its maximum at a spring's neutral position**

*V(0)=0***.**

*V(T/4)*It can be done by two methods.

1. By integrating by displacement of a spring's end from value

**(full compression) to value**

*−L***(neutral position)**

*0*

*W*

= ∫_{[−L,0]}== ∫

^{0}_{−L}F(S)·*d*

**S =**

= ∫d= ∫

^{0}_{−L}(−k·S)·**S =**

= −k·S²/2|= −k·S²/2|

^{0}_{−L}= k·L²/22. By integrating by time from

**to**

*t=0***(calculations are more complex but lead to the same result)**

*t=T/4=(π/2)·√m/k*

*W*

= ∫_{[0,T/4]}== ∫

_{0}^{T/4}F(S(t))·*d*

**S(t) =**

= ∫d= ∫

_{0}^{T/4}F(S(t))·(**S(t)/**d**t)·**d**t =**

= ∫d= ∫

_{0}^{T/4}F(S(t))·V(t)·**t =**

= ∫

·L·√k/m·sin(t·√k/m)·d= ∫

_{0}^{T/4}k·L·cos(t·√k/m)··L·√k/m·sin(t·√k/m)·

**t =**

= ∫

·sin(t·√k/m)·d= ∫

_{0}^{T/4}k·L²·cos(t·√k/m)··sin(t·√k/m)·

**(t·√k/m) =**

= ∫

·d= ∫

_{0}^{T/4}k·L²·sin(t·√k/m)··

**(sin(t·√k/m)) =**

= k·L²sin²((1/2)π√m/k·√k/m)/2= k·L²sin²((1/2)π√m/k·√k/m)/2

After cancellation of factors and taking into account that

**we obtain the same expression for work:**

*sin(π/2)=1*

*W*_{[0,T/4]}= k·L²/2What we have proven now is that the same amount of work done to compress a spring (

**) is performed by a spring when it returns to its neutral position.**

*W=k·L²/2*That's why we say that the spring accumulates

*potential energy*, when it's compressed, and releases it, when it returns back to neutral position.

Now, let's see where this energy, released by a spring during its return to a neutral position, goes. Our assumption is that it is transformed into kinetic energy of the object attached to its free end. Let's check it out.

*E*

= m·L²·(k/m)·sin²(π/2)/2 =

= k·L²/2_{kin}= m·V²(T/4)/2 == m·L²·(k/m)·sin²(π/2)/2 =

= k·L²/2

So, kinetic energy of the object, when a spring is in the neutral position, exactly equals to a spring's potential energy when it's fully compressed.

Compressing a spring results in transferring some external energy to its potential energy. Then, as the spring returns to its neutral position, it releases this potential energy into a kinetic energy of the object attached to its end. At the neutral position the potential energy of a spring is zero, but kinetic energy of the object equals to potential energy of a spring in a compressed position.

Our last step is to calculate

*full energy*of the system (

*potential energy*of a spring and

*kinetic energy*of the object attached to it) at any moment of time

**.**

*t*

*E*

= k·L²·cos²(t·√k/m)/2_{pot}(t) = k·S²(t)/2 == k·L²·cos²(t·√k/m)/2

*E*

= m·L²·k/m·sin²(t·√k/m)/2_{kin}= m·V²(t)/2 == m·L²·k/m·sin²(t·√k/m)/2

*E*_{full}= E_{pot}+ E_{kin}= k·L²/2(since

*sin²(φ)+cos²(φ)=1*)

As we see, the

*full energy*remains constant, it does not depend on the attached object's mass, only on the properties of the spring - coefficient of

*elasticity*

**and its initial displacement from the neutral position.**

*k*
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