10 points] Consider a 20 kg block attached to a spring on the (frictionless) a-a

Question

10 points] Consider a 20 kg block attached to a spring on the (frictionless) a-axis (measured in meters) with a resting position of r o. Suppose that the spring constant of the spring is 80 Newtons per meter; that is, the force required to hold the spring at a position of r is 80r Newtons in the positive r-direction. What this states is that the spring exerts a force of F(x) - -80r Newtons in the positive z-direction that counteracts the stretching force (note that negative force in the positive a-direction is positive force in the negative r-direction). If a is the acceleration (in meters per second) corresponding to the force exerted by the spring, then F(x) 20a as well. Suppose that we consider stretching the spring to a position of r units, and consider its motion upon release. Let t be the time in seconds since the release of the spring. Then a is the second derivative of z with respect to t, so F) Since F(x) =-80r as well, we have 2080r. Show that r -f(t) -sin(2t) is a solution to this differential equation, and explain why it makes sense in the context of this problem. Explain what the motion of the spring looks like, given this equation.

Explanation / Answer

20 d2x/dt2 = -80x

=> d2x/dt2 = -4x

x = sin(2t)

dx/dt = 2cos(2t)

d2x/dt2 = d/dt (2cos(2t)) = -4 sin(2t) = -4x

Hence verified

In this case , it makes sense as sin(2t) is periodic and hence it repeats itself after a fixed period (Time Period )

This is an example of Simple harmonic motion which is of the form

d2x/dt2 = -w2x where w is the angular frequency

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