How to integrate d2x/dt2 with respect to x?

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  • Thread starter Max Loo Pin Mok
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This means that the derivative of ##v^2## with respect to ##x## is ##2v = 2(dx/dt)##In summary, the conversation in chapter 3 of Vibrations and Waves by French discusses the equations of motion of a mass-spring system, including equations (3-1) and (3-2). The question at hand is how to understand the integration of the first equation with respect to displacement x. The solution involves applying the chain rule and calculating the derivative of a function with respect to another function.
  • #1
TL;DR Summary
About the mass-spring system.
Can we derive one equation of motion from another?
In chapter 3 of Vibrations and Waves by French, there is a description about the equations of motion of a mass-spring system. It was written as shown in the attached picture:
Equation of Motions.png

Here, m is the mass on the spring, k is the spring constant, x is the extension of the spring, and t is time. My problem is about figuring out how equation (3-1) becomes equation (3-2)? I tried to differentiate 1/2 m(dx/dt)2:
equation 2.png

I got an extra dt/dx which is not there in equation (3-1). Is my working wrong or is there another way of understanding what is in the textbook: "The second is, of course, the result of integrating the first with respect to the displacement x ..."?
 
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  • #2
You are not calculating correctly the derivative $$\frac{d}{dt}\frac{1}{2}m(\frac{dx}{dt})^2$$

if ##f## is a function of ##t##, what is the first derivative of ##f^2##? (w.r.t to ##t##). Then apply this for ##f=\frac{dx}{dt}##
 
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  • #3
According to the chain rule, you have
$$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx},$$ where ##v = dx/dt##.
 

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