Horizontal spring oscillations

In summary: F=ma instead of a=-xω^2?Yes, I tried that, and it gave me a value of 294m/s^2. Is that close to the value you got for number 2 with a=2d/t^2?
  • #1
fezmaster
5
0
Hey everyone. I have 2 relatively basic questions about horizontal springs. I feel like the questions are actually very simple (it's just high school physics) but I think I'm approaching it the wrong way. Any help would be greatly appreciated.
1. I'm supposed to find the spring constant when a 5.5kg mass is vibrating at the end of a horizontal spring. It reaches a maximum speed of 7.2m/s and has a maximum displacement of 0.23m. Ignore friction.
2. I'm supposed to find the acceleration of a 4.97kg mass when the displacement of the mass is 2.56m. It oscillating on the end of a horizontal spring with a frequency of 0.467s to the left.

1.
I thought I should first find the acceleration, then force (F=ma) and then solve for the spring constant (k=-F/x). To find the acceleration, I used a=V^2/2d. However, that gave me almost 113m/s. Surely that isn't correct... I feel like I should be using equations for energy somewhere, but I'm not sure which ones.
2.
I tried a similar thing here, using a=2d/t^2. I assumed that I should divide time by 2, because that's the time for a complete oscillation, and I'm only calculating half (2.56m and back to equilibrium). However, this gives me an acceleration of almost 94 m/s^2, which also doesn't feel right.

Thanks in advance!
 
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  • #2
fezmaster said:
Hey everyone. I have 2 relatively basic questions about horizontal springs. I feel like the questions are actually very simple (it's just high school physics) but I think I'm approaching it the wrong way. Any help would be greatly appreciated.
1. I'm supposed to find the spring constant when a 5.5kg mass is vibrating at the end of a horizontal spring. It reaches a maximum speed of 7.2m/s and has a maximum displacement of 0.23m. Ignore friction.
2. I'm supposed to find the acceleration of a 4.97kg mass when the displacement of the mass is 2.56m. It oscillating on the end of a horizontal spring with a frequency of 0.467s to the left.


1.
I thought I should first find the acceleration, then force (F=ma) and then solve for the spring constant (k=-F/x). To find the acceleration, I used a=V^2/2d. However, that gave me almost 113m/s. Surely that isn't correct... I feel like I should be using equations for energy somewhere, but I'm not sure which ones.


a=V2/(2d) is a relation valid for for a motion with constant acceleration. The problem is about a vibrating body, performing simple harmonic motion. The displacement is x= A sin(ωt) where A is the maximum displacement, and ω is 2pi times the frequency. You certainly have learned the formulas also for the velocity and acceleration.

fezmaster said:
2.
I tried a similar thing here, using a=2d/t^2. I assumed that I should divide time by 2, because that's the time for a complete oscillation, and I'm only calculating half (2.56m and back to equilibrium). However, this gives me an acceleration of almost 94 m/s^2, which also doesn't feel right.

Thanks in advance!

the same problem again: It is simple harmonic motion, use the appropriate formulas.

ehild
 
  • #3
Thanks, although I'm still somewhat confused. Part of that may be because we have not been taught this yet. Unfortunately, we're still expected to do the homework.
I see your equation, but I don't quite understand how it fits into my problems.

I looked up formulas for acceleration and found a=-xω^2. However, when I use that formula for number 2, I get 463m/s^2. That's an even larger number than before.


I tried something different for number 1. I used Et=1/2mvmax^2 and Et=1/2kA^2 to get 1/2mvmax^2=1/2kA^2. I rearranged to solve for k and got 5.4E3N/m. Does that work?
 
  • #4
fezmaster said:
Thanks, although I'm still somewhat confused. Part of that may be because we have not been taught this yet. Unfortunately, we're still expected to do the homework.
I see your equation, but I don't quite understand how it fits into my problems.

I looked up formulas for acceleration and found a=-xω^2. However, when I use that formula for number 2, I get 463m/s^2. That's an even larger number than before.


I tried something different for number 1. I used Et=1/2mvmax^2 and Et=1/2kA^2 to get 1/2mvmax^2=1/2kA^2. I rearranged to solve for k and got 5.4E3N/m. Does that work?

Don't just use equations, understand why you are using them.
What you did here is saying that the potential energy of the spring at maximum displacement is equal to the energy of the body at maximum velocity. Why is that true?

Regarding the formulas you are using for #2
Since you need to find the acceleration, think about what laws you can use to find out the acceleration, and then figure out what is missing from the equation you will have for the acceleration in order to solve it, using what you were given.
A small hint - try to think how you can use the frequency of oscillation to find K.

Have fun.
 
  • #5
fezmaster said:
Thanks, although I'm still somewhat confused. Part of that may be because we have not been taught this yet. Unfortunately, we're still expected to do the homework.
I see your equation, but I don't quite understand how it fits into my problems.

I looked up formulas for acceleration and found a=-xω^2. However, when I use that formula for number 2, I get 463m/s^2. That's an even larger number than before.

The equation a=-xω2 is correct, it comes from the formula for the spring force : F=-kx =ma, and from the equation for the angular frequency, ω2=k/m. x=2.56 m, the frequency is f=0.467 1/s, the angular frequency is ω=2∏f. Just plug-in.

fezmaster said:
I tried something different for number 1. I used Et=1/2mvmax^2 and Et=1/2kA^2 to get 1/2mvmax^2=1/2kA^2. I rearranged to solve for k and got 5.4E3N/m. Does that work?

That is OK.
 

What is a horizontal spring oscillation?

A horizontal spring oscillation is a type of motion in which a spring is stretched or compressed horizontally and then released, causing it to vibrate back and forth. This type of oscillation is governed by Hooke's law, which states that the force exerted by the spring is directly proportional to the distance it is stretched or compressed from its equilibrium position.

What factors affect the frequency of a horizontal spring oscillation?

The frequency of a horizontal spring oscillation is affected by three main factors: the mass of the object attached to the spring, the stiffness of the spring, and the amplitude of the oscillation. As the mass increases or the stiffness decreases, the frequency decreases. Similarly, as the amplitude increases, the frequency also increases.

How does the amplitude affect the period of a horizontal spring oscillation?

The period of a horizontal spring oscillation is the time it takes for one complete cycle of motion. The amplitude of the oscillation does not affect the period, as long as the amplitude is small. However, for larger amplitudes, the period will increase due to the non-linear relationship between force and displacement in Hooke's law.

What is the relationship between the spring constant and the period of a horizontal spring oscillation?

The spring constant, also known as the stiffness constant, is a measure of how stiff the spring is. It is directly proportional to the frequency of the oscillation and inversely proportional to the period. This means that a higher spring constant will result in a higher frequency and a shorter period, while a lower spring constant will result in a lower frequency and a longer period.

How is the energy of a horizontal spring oscillation related to its amplitude?

The energy of a horizontal spring oscillation is directly proportional to the square of its amplitude. This means that as the amplitude increases, the energy of the oscillation also increases. This is because the potential energy stored in the spring increases as it is stretched or compressed further from its equilibrium position.

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