A 8 pound object is suspended from a spring. The spring stretches an extra 4 inc

Question

A 8 pound object is suspended from a spring. The spring stretches an extra 4 inches with the weight attached. If the spring is stretched by 2 inches and released with an downward velocity of 39.1918 feet per second, find a formula describing the displacement of the object using only a single cosine function. What is the differential equation for this spring-and-mass svstem? (Use u as the dependent variable.) (14 )u''+24u=0 correct What is the formula for the displacement of the object? a(t) = | 4*(cos(sqrt(96)*tartan(24))) incorrect

Explanation / Answer

2] Let it be of form,

u = A cos (wt + phi) and speed v = -Aw sin(wt+phi)

w = sqrt (24*4) = sqrt(96) = 9.798 rad/s

Now at t=0, -2/12 = A cos phi

and -39.1918 = -A*9.798 sin phi

- 9.798 tan phi = 39.1918/(2/12) = 235.15

phi = arctan(235.15/9.798) = 1.6124 rad, A = (-2/12)/ cos 1.6124 = 4

So u = 4 cos (9.798t + 1.6124)

u is in feet, for inches multiply by 12.

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