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The total energy of a mass-spring system is the sum of its kinetic and potential energy: \(E=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2} .\) Assuming \(E\) remains constant, differentiate both sides of this expression with respect to time and show that Equation 13.3 results. (Hint: Remember that \(v=d x / d t .)\)

Short Answer

Expert verified
After differentiation and substituting the velocity definition, we get \(0 = m \frac{d^{2} x}{d t^{2}} + k x\). This is a standard expression for a harmonic oscillator, which is assumed to be Equation 13.3.

Step by step solution

01

Write down the expression for total energy

Given the total energy \(E\) of the system is equal to the sum of kinetic and potential energies and it remains constant, this gives the equation \(E=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2}\).
02

Differentiate both sides of the energy equation

Differentiating both sides with respect to time, we get \(\frac{d E}{d t} = m v \frac{d v}{d t} + k x \frac{d x}{d t}\). Since \(E\) is a constant, \(\frac{d E}{d t} = 0\). Therefore, we get \(0 = m v \frac{d v}{d t} + k x \frac{d x}{d t}\).
03

Use the definition of velocity

As noted in the hint, the velocity \(v\) is defined as \(v = \frac{d x}{d t}\). Substituting this into the obtained equation yields \(0 = m (\frac{d x}{d t}) \frac{d (\frac{d x}{d t})}{d t} + k x \frac{d x}{d t}\), simplifying gives \(0 = m \frac{d^{2} x}{d t^{2}} + k x\).

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Most popular questions from this chapter

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