# Simple Harmonic Motion of a Spring Elevator

• UrbanXrisis
In summary, the question is asking about the amplitude of the block's displacement after it has been accelerated by an elevator.
UrbanXrisis
the question is http://home.earthlink.net/~urban-xrisis/phy002.jpg

a.
$$\omega = \sqrt{\frac{k}{m}}$$
$$\omega = \sqrt{\frac{500}{2}}$$

b.
$$F=kx$$
$$\frac{g}{3} * 2kg=500x$$
$$x=1.31cm$$

c.
since the block displaces 1.31 cm then that is it's amplitude

I don't undersatnd what they mean by inital phase angle. The initial phase would be the mass at -1.31cm moving upwards. I would just be...
$$x=- \omega^2 A cos(\omega t + \phi)$$
$$\phi=0$$

or... I think this might be a valid answer...

$$x_i=Acos\phi$$
$$-1.31=1.31cos\phi$$
$$\phi = \pi$$

am I doing this all correctly?

Last edited by a moderator:
Looks good to me. (Yes, $\phi = \pi$.)

Could you explain to me what you did in part b. please?

When the car is accelerating, the spring must exert an additional force equal to $ma = mg/3$, thus the spring must stretch an additional $x = mg/(3k)$.

But isn't this just the new equilibrium position of the block? After this new position has been attained, the sum of the force of the block is zero but it will have picked a speed in going from rest to this point while accelerating according to

$$a = -kx/m + g/3$$

Therefor it will oscilate around $x = mg/(3k)$.

With no acceleration, the equilibrium point will be where the spring is stretched an amount $mg/k$; when the car is accelerating at $g/3$, the equilibrium position will be at a stretch of $mg/k + mg/(3k)$.

The initial condition is that the car is accelerating, thus the initial position of the mass is at $x_0 = -mg/k -mg/(3k)$. At t=0, the acceleration stops (but the speed remains at whatever it is). So, at t=0 the equilbrium position is $mg/k$, thus the mass is displaced $mg/(3k)$ below that equilibrium point. It will oscillate about the equilibrium point with an amplitude of $mg/(3k)$.

Oops, I had misread the question. I tought the block was initially at rest, THEN the elevator started accelerating, THEN it sudently stopped. Sowwy.

## What is the "Spring Elevator Question"?

The "Spring Elevator Question" is a thought experiment that explores the concept of energy conservation in a mechanical system. It involves a spring attached to the ceiling of an elevator, with a block sitting on top of the spring. When the elevator is at rest, the spring is compressed by a certain distance. The question is, what happens to the block and the spring when the elevator begins to move?

## Why is the "Spring Elevator Question" important?

The "Spring Elevator Question" is important because it demonstrates the principle of conservation of energy, which is a fundamental concept in physics. It also helps to illustrate the relationship between potential energy and kinetic energy in a mechanical system.

## What is the answer to the "Spring Elevator Question"?

The answer to the "Spring Elevator Question" is that the block will remain stationary relative to the elevator, while the spring will expand or contract depending on the direction of the elevator's motion. This is because the potential energy stored in the compressed spring is converted into kinetic energy as the elevator moves.

## How does the "Spring Elevator Question" relate to real-life situations?

The "Spring Elevator Question" can be applied to real-life situations, such as a car driving over a speed bump or a person jumping on a trampoline. In both cases, the car or person experiences a change in potential energy, which is then converted into kinetic energy.

## What are some variations of the "Spring Elevator Question"?

Some variations of the "Spring Elevator Question" involve changing the starting position of the elevator, the mass of the block, or the strength of the spring. These variations can help to further explore the relationship between potential and kinetic energy in a mechanical system.

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