Spring and Elevator Questions i am stuck

In summary: If you are calling up, then you would use a minus sign, and if you are calling down, then you would not.
  • #1
hemetite
50
0
A particle of mass 0.2kg hangs from and ideal spring. In equilibrium, the spring is stretched by amount of 0.49cm. The spring is then suspended from the ceiling of a lift and hangs motionless relative to the lift as the lift descends with a constant velocity of 2.0m/s. The lift then suddenly stops.

a) With what amplitude will the particle oscillate?
b) What is the equation of the motion for the particles?

Take the upward directions to be positive and g= 9.8ms-2
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solution for question a)

I know that at the point of the elevator stops. Vmax= 2.0m/s
at equilibirium
Fg=-kx
0.2*9.8= -(k)(0.0049)
k=-400
w=-sqrt(400/0.2)= -44.72

using v(t)= -wA sin(wt + teta)
at equilibirium teta= 0, v(t)=Vmax

2=-44.72*A sin (wt)

A= -(2/(44.72*sin(wt))

I am stuck now...suddenly my brain jammed...what is the value of the t here?
t= 0 or t= 2?

or since it is at equilibrium when at the point release. therefore

Vm= wA
Which then,
A= -(2/44.72)= 0.044722719
?

Please help...
 
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  • #2
I believe that the second part you wrote is correct. 4.4 cm seems like a reasonable amplitude for this problem, as well, since the spring is strong but the elevator was moving relatively fast.

The first way you tried would also work. The block/spring system is still in equilibrium when the elevator stops, so you can call that t=0. But its first maximum amplitude doesn't occur until the sin() term is at a maximum, which means wt = pi/2 and your equation simplifies to what you wrote later since sin(pi/2) is one. You don't really need to find t, but you could if you wanted.
 
  • #3
Thanks..i have to agree with you on the that hight amplitude for sin(pi/2) whereas cos (0)

The answer should be correct in my rational thinking...:)

thank you again...
 
  • #4
for the second part of the questions...
b) What is the equation of the motion for the particles?

"equation of the motion for the particles" in another word...the equation for the displacement for the particles..since the particle is oscillating.

Am i right to say i will be using the displacement equations.

x(t)= A cos (wt + teta)

i konw that (wt + teta) is the initial phase of the particle oscillating...but i dun think is right for me to put pi/2 as the teta...i believe that this is right coz...i know that the value of teta depends on the displacement and the velocity at t=0. therefor that teta is fixed constant value...


putting the all the values in will get

x(t)= 0.044 cos (44.721 t + pi/2)

should i put back the minus sign??

am i right or wrong here?

is this the equation?


i stuck with my thoughts here...?
 
Last edited:
  • #5
What you wrote is correct, I think, but you can also re-write the expression with a sin() term, since [tex]cos{\left(\theta + \pi/2\right)} = sin{\theta}[/tex].

Some people like using sin() because it shows more clearly that the position is 0 at t=0. cos() is used often because most of the time when learning SHM you are talking about a spring which is displaced and then released from rest, which is not the case here.

Whether you put in a minus sign or not depends on which direction (up/down) you are calling positive.
 

1. What is the relationship between spring and an elevator?

The spring in an elevator serves to counteract the weight of the elevator car and its passengers, allowing the elevator to move smoothly and efficiently between floors.

2. How does a spring in an elevator work?

The spring is typically compressed when the elevator car is heavier, such as when passengers enter or exit, and expands when the car is lighter. This allows the elevator to maintain a consistent weight and move up and down smoothly.

3. What happens if the spring in an elevator breaks?

If the spring in an elevator breaks, the elevator car will become significantly heavier and may not be able to move. This can cause the elevator to become stuck between floors until the spring is repaired or replaced.

4. How often do springs in elevators need to be replaced?

The lifespan of a spring in an elevator can vary depending on factors such as usage and maintenance. However, most springs are designed to last for several years before needing to be replaced.

5. Are there different types of springs used in elevators?

Yes, there are several types of springs that can be used in elevators, including helical springs, coil springs, and leaf springs. Each type has its own unique characteristics and is chosen based on the specific needs of the elevator system.

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