Integrating a second derivative-involving solution for Simple Harmonic Motion

In summary, the conversation is about the process of integrating m(d^2x/dt^2)=-kx to obtain the general solution of x=Acos(wt+c). The difficulty lies in integrating the second derivative of acceleration, but the conversation suggests rearranging it as shown and then integrating it twice. The speaker asks for help in dealing with the terms involving v^2, x^2, and -k/m. The expert suggests finding a value for C from the initial conditions and then converting v to dx/dt and integrating again to obtain the general solution.
  • #1
insertnamehere
50
0
Hello, i am now in the process of integrating m(d^2x/dt^2)=-kx which i know i will have to do twice in order to obtain the general solution to simple harmonic motion, x= Acos(wt+c) c=phi

but I'm just having problems with the second derivative of acceleration (d^2*x/dt^2) when it comes to integrating, I tried separating them into dv/dt and then dv/dx and dx/dt, so i obtain v, etc... but this complicates things even more for me! It would be so great if someone could give me a hint as to where i can start in integrating with this second derivative...
Thank you very much for your time.
 
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  • #2
insertnamehere said:
Hello, i am now in the process of integrating m(d^2x/dt^2)=-kx which i know i will have to do twice in order to obtain the general solution to simple harmonic motion, x= Acos(wt+c) c=phi

but I'm just having problems with the second derivative of acceleration (d^2*x/dt^2) when it comes to integrating, I tried separating them into dv/dt and then dv/dx and dx/dt, so i obtain v, etc... but this complicates things even more for me! It would be so great if someone could give me a hint as to where i can start in integrating with this second derivative...
Thank you very much for your time.
The 2nd differential is often rearranged like this,

[tex]\frac{d^2x}{dt^2}= v\frac{dv}{dx}[/tex]

and you get that this way,

[tex]\frac{d^2x}{dt^2}=\frac{d}{dt}(dx/dt)=\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= \frac{dv}{dx}v = v\frac{dv}{dx}[/tex]

substitute for [tex]\frac{d^2x}{dt^2}[/tex] to [tex]v\frac{dv}{dx}[/tex] then do the integration.
 
  • #3
Yes, i tried doing that, but then eventually i get stuck with
(v^2)/2= (-k/m)(x^2)/2 + C
Would i have to integrate again to obtain the general solution of
x=Acos(wt+c)? What can i do with the v^2 and the x^2, and the negative k/m?
 
  • #4
Your halfway there.
From the initial conditions of the problem, work out a value for C.
rewrite the eqn as v = whatever ...
convert v to dx/dt and integrate agan.
 

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object oscillates back and forth around a central position due to the restoring force of a linear system, such as a spring or pendulum. It is characterized by a sinusoidal motion with a constant amplitude and frequency.

2. How is a second derivative involved in SHM?

In SHM, the second derivative of the position function with respect to time represents the acceleration of the object. This acceleration is directly proportional to the displacement of the object from its equilibrium point, and is directed towards the equilibrium point. This is known as Hooke's Law.

3. What is the equation for SHM and how does it involve the second derivative?

The equation for SHM is x(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. The second derivative of this equation is -A*ω^2*cos(ωt + φ), which represents the acceleration of the object at any given time.

4. How do you integrate a second derivative in SHM?

To integrate the second derivative in SHM, you can use the equation x(t) = -(1/ω^2)*d^2x/dt^2. This will give you the position function as a function of time. You can then solve for the constants A and φ using initial conditions, such as the initial position and velocity of the object.

5. What are the applications of SHM in the real world?

SHM has many real-world applications, including in the study of waves, sound, and electricity. It is also used in engineering for the design of structures with minimal oscillation, such as suspension bridges and tall buildings. SHM can also be seen in everyday objects, such as a swinging pendulum clock or a mass on a spring.

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