A coil spring is such that a 25 lb weight would stretch it 6 inches. The spring

Question

A coil spring is such that a 25 lb weight would stretch it 6 inches. The spring is suspended from the ceiling, a 16 lb weight is attached to the end, and the weight then comes to rest in its equilibrium position. It is then pulled down 4 inches from its equilibrium posistion and released at t=0 with an initial velocity of 2ft/s, directed upwards. Assume there is no damping force and no force due to gravity.

a. Determine the resulting displacement of the weight as a function of time.

b. Find the amplitude, period, and frequency of the resulting motion.

c. At what time does the weight first pass through its equilibrium posistion and what is its velocity at this instant?

Explanation / Answer

Spring stiffness k = Force/displacement = 25/(6/12) = 50 lb/ft

Newton's second law: mx'' = -kx or mx'' + kx = 0

Putting x = a Sin (wt + phi) we get

x' = aw Cos (wt + phi)

and x'' = -aw^2 Sin (wt + phi) = -w^2 x

Thus, -mw^2 x + kx = 0

This gives Natural frequency w = sqrt(k/m) = sqrt(50*32.2 / 16) = 10 rad/s

At t = 0, we have displacement x = 0 and velocity x' = -2 ft/s (negative since upwards)

Hence, 0 = a Sin phi and,

-2 = -a*10 Cos (phi)

Solving both these,

a = 0.2 ft and phi = 0

Thus, x = 0.2 Sin (10*t)

Amplitude = a = 0.2 ft,

Frequency = w = 10 rad/s = 10/2pi Hz = 1.592 Hz

Time period = 2pi/w = 0.628 s

At equilib. pos. , x = 0

So, 10*t = pi

t = 0.314 s

Vel. at equil. pos = aw = 10*0.2 = 2 ft/s

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