# Hooke's Law Second Order Differential Equation

• Kernul
In summary: Am I right?Sorry, I just want to be sure to understand so I'm going to write all.We know that the standard form is ##ax''(t) + bx'(t) + cx(t) = 0##.Then we know that:$$x''(t) = \frac{d^2 x}{d t^2} = -\frac{k}{m} x(t)$$$$x'(t) = \frac{d x}{d t} = -\frac{k}{m} \int x(t) d t$$Substituting these into the standard form we get:$$-\frac{k}{m} a x(t) -\frac{k}{m} b \ Kernul ## Homework Statement A mass ##m## on a frictionless table is connected to a spring with spring constant ##k## so that the force on it is ##F_x = -kx## where ##x## is the distance of the mass from its equilibrium position. It is then pulled so that the spring is stretched by a distance ##x## from its equilibrium position and at ##t = 0## is released. Write Newton’s Second Law and solve for the acceleration. Solve for the acceleration and write the result as a second order, homogeneous differential equation of motion for this system. ## Homework Equations Newtons's Second Law. Hooke's Law Differential Equations ## The Attempt at a Solution I write the the Newton's Second Law and solve for the acceleration:$$F = m a_x = - k xa_x = -\frac{k}{m} x$$Now it tells me to write the result as a second order, homogeneous differential equation of motion. I don't quite get how I should do this but I think this way: I write ##a_x = -\frac{k}{m} x## as ##\frac{d v_x}{d t} = -\frac{k}{m} x## and multiplying both sides for ##d t## and integrating I have:$$v_x(t) = -\frac{k}{m} x t + C$$where ##C## is a constant and would actually be ##v_{0x} = 0## Same thing again with ##\frac{d x}{d t} = -\frac{k}{m} x t## and having:$$x(t) = -\frac{k}{2m} x t^2 + C$$Now, should I put all this like$$x''(t) + x'(t) + x(t) = 0$$and so$$(-\frac{k}{m} x) + (-\frac{k}{m} x t) + (-\frac{k}{2m} x t^2) =0$$Is this the correct way? Kernul said: I write the the Newton's Second Law and solve for the acceleration:$$F = m a_x = - k xa_x = -\frac{k}{m} x$$Now it tells me to write the result as a second order, homogeneous differential equation of motion. I don't quite get how I should do this Can you express ##a_x## as a second derivative? I write ##a_x = -\frac{k}{m} x## as ##\frac{d v_x}{d t} = -\frac{k}{m} x## and multiplying both sides for ##d t## and integrating I have:$$v_x(t) = -\frac{k}{m} x t + C$$Here you treated ##x## as a constant. But ##x## is a function of time. You are asked to find a second-order, homogeneous differential equation such that if you solved it, it would give you the answer for ##x(t)##. But the statement of the question does not ask you to solve the differential equation. TSny said: Can you express ##a_x## as a second derivative? Here you treated ##x## as a constant. But ##x## is a function of time. You are asked to find a second-order, homogeneous differential equation such that if you solved it, it would give you the answer for ##x(t)##. But the statement of the question does not ask you to solve the differential equation. I can express ##a_x## as ##\frac{d^2 x}{d t^2}##. And yes, you're right, I wrongly treated it as a constant. So I should have something like ##v_x = -\frac{k}{m} \int x(t) dt##. Kernul said: I can express ##a_x## as ##\frac{d^2 x}{d t^2}##. Yes. And yes, you're right, I wrongly treated it as a constant. So I should have something like ##v_x = -\frac{k}{m} \int x(t) dt##. A differential equation of motion for the mass would be a differential equation involving time derivatives of the unknown function ##x(t)##. No need to bring in ##v_x##. Kernul said: I can express ##a_x## as ##\frac{d^2 x}{d t^2}##. Good, now substitute that into your equation$$a_x = -\frac{k}{m} x$$What do you get? Kernul TSny said: A differential equation of motion for the mass would be a differential equation involving time derivatives of the unknown function ##x(t)##. No need to bring in ##v_x##. So I should write ##\frac{d x}{d t}## instead of ##v_x##? Chestermiller said: Good, now substitute that into your equation$$a_x = -\frac{k}{m} x$$What do you get? I would get$$\frac{d^2 x}{d t^2} = -\frac{k}{m} x$$Kernul said: So I should write ##\frac{d x}{d t}## instead of ##v_x##? I would get$$\frac{d^2 x}{d t^2} = -\frac{k}{m} x$$Yes, that's essentially it! You can put into "standard form" by arranging it in the form ##ax''(t) + bx'(t) + cx(t) = 0##. The zero on the right makes it homogenous. http://mathworld.wolfram.com/HomogeneousOrdinaryDifferentialEquation.html Kernul TSny said: Yes, that's essentially it! You can put into "standard form" by arranging it in the form ##ax''(t) + bx'(t) + cx(t) = 0##. The zero on the right makes it homogenous. http://mathworld.wolfram.com/HomogeneousOrdinaryDifferentialEquation.html But substituting I would have something like this, right:$$x''(t) + x'(t) + x(t) = 0-\frac{k}{m} x(t) -\frac{k}{m} \int x(t) dt + x(t) = 0x(t) = \frac{k}{m} x(t) + \frac{k}{m} \int x(t) dt$$Kernul said: But substituting I would have something like this, right:$$x''(t) + x'(t) + x(t) = 0$$No. Why are you writing this equation? You have already essentially answered the question. Can you put your result into the standard form ##ax'' + bx' + cx = 0##, where ##a, b## and ##c## are constants? TSny said: No. Why are you writing this equation? You have already essentially answered the question. Can you put your result into the standard form ##ax'' + bx' + cx = 0##, where ##a, b## and ##c## are constants? Sorry, I just want to be sure to understand so I'm going to write all. We know that the standard form is ##ax''(t) + bx'(t) + cx(t) = 0##. Then we know that:$$x''(t) = \frac{d^2 x}{d t^2} = -\frac{k}{m} x(t)x'(t) = \frac{d x}{d t} = -\frac{k}{m} \int x(t) d t$$Substituting these into the standard form we get:$$-\frac{k}{m} a x(t) -\frac{k}{m} b \int x(t) dt + c x(t) = 0$$Am I right? Kernul said: Sorry, I just want to be sure to understand so I'm going to write all. We know that the standard form is ##ax''(t) + bx'(t) + cx(t) = 0##. Then we know that:$$x''(t) = \frac{d^2 x}{d t^2} = -\frac{k}{m} x(t)x'(t) = \frac{d x}{d t} = -\frac{k}{m} \int x(t) d t$$No need to write the second equation. Rearrange the first equation to get 0 on the right and you're done. TSny said: No need to write the second equation. Rearrange the first equation to get 0 on the right and you're done. So simply this?$$x''(t) + \frac{k}{m}x(t) = 0

Yes. Writing the answer as ##x''(t) = -\frac{k}{m}x(t)## is probably OK, too. Writing it with 0 on the right just puts the differential equation in standard form. The question statement is not clear on whether or not you need to put it into standard form. Since it says, "solve for the acceleration" it could be that they actually want ##x''(t) = -\frac{k}{m}x(t)##. Who knows.

Kernul
TSny said:
Yes. Writing the answer as ##x''(t) = -\frac{k}{m}x(t)## is probably OK, too. Writing it with 0 on the right just puts the differential equation in standard form. The question statement is not clear on whether or not you need to put it into standard form.
Oh, I get it now. I think I got confused before.
Yeah, well, I'll use the standard form. Thank you very much!

OK. Good work.

## 1. What is Hooke's Law Second Order Differential Equation?

Hooke's Law Second Order Differential Equation is a mathematical equation that describes the relationship between the force applied to an elastic material and its resulting displacement. It is based on the principle that the force applied is directly proportional to the displacement of the material.

## 2. What is the formula for Hooke's Law Second Order Differential Equation?

The formula for Hooke's Law Second Order Differential Equation is F = -kx, where F is the force applied, k is the spring constant, and x is the displacement of the material.

## 3. How is Hooke's Law Second Order Differential Equation applied in real life?

Hooke's Law Second Order Differential Equation is commonly used in engineering and physics to analyze the behavior of elastic materials, such as springs and rubber bands. It is also used in the design of structures and machines to ensure they can withstand the forces applied to them.

## 4. What is the significance of the second order in Hooke's Law Second Order Differential Equation?

The second order in Hooke's Law Second Order Differential Equation represents the fact that the relationship between force and displacement is not linear, but rather it is quadratic. This means that the force is directly proportional to the square of the displacement.

## 5. How does Hooke's Law Second Order Differential Equation relate to Newton's Second Law of Motion?

Hooke's Law Second Order Differential Equation is closely related to Newton's Second Law of Motion, which states that the force applied to an object is equal to its mass multiplied by its acceleration. In the case of Hooke's Law, the acceleration is replaced by the second derivative of displacement, resulting in the same equation: F = ma = m(d^2x/dt^2).

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