How Does Elevator Motion Affect Spring Oscillation?

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SUMMARY

The discussion focuses on the effects of elevator motion on spring oscillation, specifically analyzing a 2.0 kg mass attached to a spring with a spring constant of 500 N/m. When the elevator is stationary, the spring stretches by 0.039 m. Under a constant upward acceleration of 1/3 g, the spring stretches to 0.052 m. Upon stopping, the mass oscillates with an amplitude of 0.01306 m, a frequency of 16.58 rad/sec, and a phase of π/2. The calculations utilize Newton's second law and harmonic motion principles.

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[SOLVED] Spring Oscillation Problem!

1. A 2.0 kg mass hangs at rest from a harmonic spring with a spring constant of 500 N/m inside an elevator that is stationary.
a) by how much is the spring stretched
b) suppose that the elevator is rising with a constant upward accel of 1/3 g. By how much is the spring stretched now.
c) The elevator stops at t=0. You observe that the mass starts to oscillate. What are the ampliturde A, frequency f, and the phase phi of the oscillation.




2. F=ma=kx-mg, phi= inverse tan[ wx(o)/v(0)].



3. I think i got a and b but am confused with c.
a) F=kx-mg.
mg=kx = (2)(9,8)= (500)x
x= 0.039 m
b) ma=kx-mg
(2)(9.8/3)=(500x)-2(9,8)
x=0.052 m.
c) w= sqrt(k/M)= 16.58 rad/sec
x(0)=0.052-0.039=0.01306. I think this is x(0) but what is v(o)?
 
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clipperdude21 said:
3. I think i got a and b but am confused with c.

a)F=kx-mg.
mg=kx = (2)(9,8)= (500)x
x= 0.039 m

OK.

b) ma=kx-mg
(2)(9.8/3)=(500x)-2(9,8)
x=0.052 m.
OK.

c) w= sqrt(k/M)= 16.58 rad/sec
x(0)=0.052-0.039=0.01306. I think this is x(0) but what is v(o)?
A = 0.052-0.039=0.01306. (Edited.)

You can find v0 from x0, m, k etc, or not? Any other idea comes to mind?
 
Last edited:
after thinking about it, i think its easier than i thought. we calculated w. To find the amplitude of the oscillation we take the answer in b and subtract it from a. The phase is then easy to calculate since x(o)=Asin(phi) and if x(o)=amplitude, then sin(phi)=1 and the phase is pi/2
 
I have edited the answer for (c), because I misunderstood the meaning of x(0). The amplitude in (c) is 0.052-0.039=0.01306.

The rest you know.
 

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