Simple Harmonic Motion & a vertical spring

In summary, a block is dropped from a height of 3.0 cm onto a spring standing upright on a table. The block sticks to the top of the spring and oscillates with an amplitude of 10 cm. The problem is to determine the oscillation frequency. One approach was to use energy equations, but this led to an incorrect answer due to an incorrect assumption about the velocity. The correct approach involves finding the equilibrium point and using the energy conservation equation at that point and at the beginning point. By relating the frequency to the spring constant and using other known equations, the correct frequency of 1.83 Hz can be found.
  • #1
danielle36
29
0

Homework Statement



A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height of 3.0 cm above the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?
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My first thought was I should be applying energy equations to solve...

[tex] U_{g_{i}} = K_{f} [/tex]
[tex] mgy = 1/2mv^{2} [/tex]
[tex] (9.81m/s^{2})(.03m) = 1/2v^{2} [/tex]
[tex] v_{max} = 0.767 m/s [/tex]

[tex] v_{max} = 2 \pi fA [/tex]
[tex] f = 1.22 Hz [/tex]


------

The answer is supposed to be 1.83 Hz..

I'm thinking I'm running into trouble because I made the assumption that the velocity just before the block hits the spring will be the same as the max velocity for the oscillation, but this is the only way I can think to do this question since I don't know the spring constant

Any help would be appreciated
 
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  • #2
Hi danielle36,

That is where your mistake is. The block does not begin slowing down as soon as it hits the spring. It will not begin slowing down until the spring force upwards is greater than the gravity force downwards.

This means the equilibrium point is some distance d below where the block first touches the spring, and that's where the maximum velocity occurs. If you follow the same basic idea that you did in your post, but choose the real equilibrium point, you can get the right answer. (To solve it I think you'll also have to use some other facts about the spring, like what defines the equilibrium point and how the frequency is related to the spring constant.) What do you get?
 
  • #3
Well I've tried a few things and I still haven't found the right method... I don't know of a way to relate the frequency and spring constant, so I tried using (mg/k)[tex]^2[/tex] as my value for v in 1/2mv[tex]^2[/tex] but it didn't leave me with anything useful... So I'm really not sure where to go from here
 
  • #4
danielle36 said:
Well I've tried a few things and I still haven't found the right method... I don't know of a way to relate the frequency and spring constant, so I tried using (mg/k)[tex]^2[/tex] as my value for v in 1/2mv[tex]^2[/tex] but it didn't leave me with anything useful... So I'm really not sure where to go from here

I would try writing the energy conservation equation at the beginning point and at the equilibrium point. You've already written down the equation relating vmax and amplitude/frequency, so you can get rid of vmax, for example, in terms of f and A.

That still leaves quite a few unknown, which you can get rid of by using other relations that you know. For example, if the equilibrium point is a distance d below where the spring starts, how is d related to other parts of the problem? Another important thing is how is the frequency related to the spring constant?

Anyways, by starting with the energy equation I think you should be able to get rid of all the unknowns except the frequency. What do you get?
 

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth between two points along a straight line. It occurs when the restoring force on an object is directly proportional to the displacement of the object from its equilibrium position.

2. How is a vertical spring used in simple harmonic motion?

A vertical spring is commonly used to demonstrate simple harmonic motion. The spring is attached to a fixed point at one end and a mass at the other end. When the mass is pulled down and released, it will oscillate up and down due to the restoring force of the spring.

3. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A*cos(ωt+φ), where x is the displacement from equilibrium, A is the amplitude (maximum displacement), ω is the angular frequency, and φ is the phase constant.

4. How does the mass affect simple harmonic motion?

The mass attached to the vertical spring affects the period and frequency of the simple harmonic motion. A heavier mass will result in a longer period and lower frequency, while a lighter mass will result in a shorter period and higher frequency.

5. Can simple harmonic motion occur without a restoring force?

No, simple harmonic motion requires a restoring force to occur. Without a restoring force, the object will not oscillate back and forth and will instead remain at rest or move with a constant velocity.

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