An air-track glider of mass 0.5 kg is attached to one end of a spring of negligi

Question

An air-track glider of mass 0.5 kg is attached to one end of a spring of negligible mass. The other end of the spring itself is fixed. The glider is then set to oscillating about the point x = 0 (the point where the spring is in its relaxed state). The spring constant is measured to be 5 N/m. First, you pull the glider until the spring is stretched by a length A as measured from its relaxed state. You then release the glider.

You now attach another spring of negligible mass to the free end of the glider. This second spring has a spring constant of 7 N/m and has one of its end fixed. You now set the glider to oscillating.

You now attach ceramic magnets on the glider in such a way as to provide some damping. You then plot the amplitude of oscillation as a function of time. You observe that, in FIVE seconds, the amplitude DECREASES by a factor of FOUR. The total mass of the glider PLUS magnets is now 0.7 kg. [Note that this differs from the lab in that in the lab we made efforts to keep the massunchanged.]

The second spring is now attached to a mechanical vibrator such as the one found in the lab. This vibrator now drives the system. You increase the driving frequency until you observe a sharp increase in the amplitude (resonance phenomenon). You adjust the frequency until the amplitude is maximal.

Explanation / Answer

a. 0.5 kA^2 = 0.5 mv^2

A^2 = 0.5*2*2/(5)

A = 0.632 m

b.f= 1/2pi)(sqrt(k/m)

f = (1/2*3/14)(sqrt(5/0.5) = 0.5 Hz

c. t = 1/f = 1/0.5 = 2 secs

2.d.Keff = K1+k2 = 5+7 = 12 N/m

e.. f = 1/2pi )(sqrt(12/0.5)

f = 0.78 Hz

f. A = xe^bt/2m

e^bt/2m = 1/4 = 0.25

bt/2m =1.38

b = 1.38 *2*0.7/5   = 0.39

g. w= sqrt( wo^2 + b^2/2m^2)

w = 0.78 Hz

h. Q = 4.1

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.