advanced math A mass weighing 32 pounds stretches a spring 2

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s. amplitude ft period s How many complete cycles will the mass have completed at the end of 3 pi seconds? Cycles

Solution

Solution :

Here, we know that m*g = 32 lbs(force), so m = 32 lbs(mass) and that the spring is extended by -2 feet by the force of 32 pounds, so:

32 pounds(force) = k*2ft

k = 16 lbs(force)/ft.

sqrt(k/m) = sqrt(16 lbs(force)/ft)/(32 lbs(mass)) = 4.0/sec = 4.0 Hz

This quantity is the angular frequency of the motion, usually denoted by .

The angular frequency is related to the period, T, by:

= 2*/T

T = 2*/

In this case, T = 2*/4Hz = /2 sec, so after 3 seconds, the mass will have oscillated through 6 complete periods.

The amplitude of the oscillations is given by magnitude of the integration constant, c, but to calculate c, we first need to know the value of .

In this problem the phase is:

= arctan((x0*k + m*g)*sqrt(k/m)/(k*v0)

= arctan((1ft*16lbf/ft + 32lbf)*(4/sec) / (16lbf/ft * 6ft/sec)

= arctan(2)

The amplitude is given by:

c = (x0*k + m*g)/(k*sin())

c = (1ft * 16lbf/ft + 32lbf)/(4/sec * sin(atan(2)))

c = 13.42 pound-force seconds