Undergrad How to integrate d2x/dt2 with respect to x?

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SUMMARY

The discussion focuses on the integration of the kinetic energy equation in a mass-spring system as described in chapter 3 of "Vibrations and Waves" by French. The user seeks clarification on how equation (3-1) transitions to equation (3-2) through differentiation and integration with respect to displacement x. The correct application of the chain rule is emphasized, specifically how to derive the first derivative of \( f^2 \) with respect to time t, where \( f = \frac{dx}{dt} \). The conclusion highlights the importance of understanding the relationship between velocity and acceleration in this context.

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  • Understanding of classical mechanics, specifically mass-spring systems.
  • Familiarity with calculus, particularly differentiation and integration.
  • Knowledge of the chain rule in calculus.
  • Basic concepts of kinetic energy in physics.
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  • Study the application of the chain rule in calculus.
  • Explore the derivation of kinetic energy equations in classical mechanics.
  • Learn about the equations of motion for mass-spring systems.
  • Investigate the relationship between velocity, acceleration, and displacement in physics.
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Students of physics, educators teaching classical mechanics, and anyone studying the mathematical foundations of motion in mass-spring systems.

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TL;DR
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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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}##
 
Last edited:
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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