Consider a mass m attached to a spring. Suppose there is no grav force acting on

Question

Consider a mass m attached to a spring. Suppose there is no grav force acting on m so that the only force is from the spring. Let the relaxed undistorted length of the spring the x0. Hooke's law says that the force actin on m is f=-k(x-x0), where k is a constant.

Upon letting E=x-x0 (note: E = Xi) be the displacement of the spring from its undistorted length x0, then

m(d^2 E/dt^2) +kE = 0 (note: d^2 E/dt^2 is the second derivative of E with respect to t)

Given that the mass starts at E=0 with an initial velocity v0, show that the displacement is given by E(t) = v0 (m/k)^0.5 sin[(k/m)^0.5].

Interpret and discuss this solution. What does the motion look like? What is the frequency? What is the amplitude?

Explanation / Answer

Just get any book on classical mechanics. This is present in ANY book on classical mechanics. My advice would be to look into the chapters of Harmonic oscillator in Feynman lectures in physics Volume 1

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