Hooke's Law and differential equations

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SUMMARY

The discussion focuses on solving the second-order differential equation derived from Hooke's Law, represented as -kx = m(d²x/dt²). The equation is rewritten as d²x/dt² + (k/m)x = 0, leading to a general solution of the form x(t) = Ccos(ωt) + Dsin(ωt), where ω = √(k/m). The transition from complex exponentials to trigonometric functions is explained using the definitions of cosine and sine in terms of exponentials. The conversation emphasizes the importance of understanding boundary conditions in determining the constants in the general solution.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with Hooke's Law and its mathematical representation
  • Knowledge of complex numbers and their exponential forms
  • Basic concepts of boundary conditions in differential equations
NEXT STEPS
  • Study methods for solving first-order differential equations
  • Learn about boundary value problems in differential equations
  • Explore the applications of Hooke's Law in physics and engineering
  • Investigate the relationship between complex exponentials and trigonometric functions
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Students of physics and mathematics, particularly those studying mechanics and differential equations, as well as educators looking for clear explanations of these concepts.

Gyroscope

Homework Statement


-kx=m\frac{d^2x}{dt^2}

I don't know how to solve differential equations, can someone show me how to do it, with this example.
 
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This is an example of a second order differential equation; let's rewrite it as \frac{d^2x}{dt^2}+\frac{k}{m}x=0 This will have a general solution of the form x=Aeat. Plug this into the equation and we obtain the equation a2+k/m=0, which gives us the values for a; namely a=\pm \sqrt{-\frac{k}{m}}=\pm i\sqrt{\frac{k}{m}}. Letting \sqrt{k/m}=\omega we obtain x(t)=Ae^{i\omega t}+Be^{-i\omega t}. Now, recognising that this is the form of a sum of cosine and sine functions, we obtain the general solution x(t)=C\cos(\omega t)+D\sin(\omega t)

I hope you follow that; however, if you have not studied differential equations, then I would suggest first learning about first order differential equations, and then moving onto second order equations (although, I suspect you want this solution for a specific purpose).

[edit: I didn't notice that this was in the homework forum, and so a full solution shouldn't be given; however, I suspect from previous posts that you may be self-learning the subject, and so this is not a homework question]
 
Last edited:
Why do you need both solutions?
 
Gyroscope said:
Why do you need both solutions?

Since both e^{i\omega t} and e^{-i\omega t} are solutions to the equation, then the general solution will be a linear combination of the two, with the constants determined by the boundary conditions.
 
Thanks cristo. How can you pass from e^(something) to cosine and sine functions?
 
Gyroscope said:
Thanks cristo. How can you pass from e^(something) to cosine and sine functions?

Use the definition of complex exponential, namely e^{\pm i\theta}=\cos\theta \pm i\sin\theta. With a bit of rearrangement, we find that \cos\theta =\frac{1}{2}(e^{i\theta}+e^{-i\theta}) and \sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})
 
Gyroscope said:

Homework Statement


-kx=m\frac{d^2x}{dt^2}

I don't know how to solve differential equations, can someone show me how to do it, with this example.

While the method described above is very useful and practical, don't forget what the equation is asking:

Can you determine which functions x(t) are proportional to minus their second derivatives?

You don't need a [crash] course in differential equations to answer that question.
 

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