# Frequency of a wire/spring system?

• Liam B
In summary, the conversation discusses a wire that is fixed on one end and under tension on the other end due to a spring. The first mode of vibration is found to be 73.6 Hz using a fixed/fixed closed form solution, but in a vibration test it is found to be 110 Hz. The speaker suspects an error in the calculation, possibly due to dropping a factor of root 2. The conversation also mentions the wire's material and diameter.
Liam B
Hi, I have a wire that is fixed on one end and is under 2 lbs (approx) of tension on the other end due to a spring. Using a fixed/fixed closed form solution I found the first mode to be 73.6 Hz. In a vibration test, we found the first mode to be 110 Hz. Does this seem reasonable and where do you think my error is coming from? The wire is made of 7x7 braided stainless steel and is .018 inches diameter.

Liam

Welcome to the forum.

Maybe I could help you more effectively if you showed your work. But I do note that the ratio of the two is suspiciously close to the square root of 2. Did you drop a factor of root 2?

Thank you, I'm happy to be here.

No, that is a good question though. The 73 Hz. is correct and I have checked it with several different sources. It assumes however that both ends are fixed and in reality one end of the wire is attached to a spring. I just didn't think that I would get 1.5x error.

My original equation was

And this gives the 73Hz.
My Setup:

This is the end with the spring. The other end is fixed.

Last edited:
And how did you measure the tension?

nasu said:
And how did you measure the tension?
I found weight that measured 2 lbs, hung it off the end of the wire which compressed the spring on the other end, and then crimped and cut the wire at that length.

## 1. What is the formula for calculating the frequency of a wire/spring system?

The formula for calculating the frequency of a wire/spring system is:
f = 1 / (2π) * √(k/m)

Where f is the frequency, k is the spring constant, and m is the mass of the object attached to the spring.

## 2. How does the length of the wire/spring affect the frequency?

The length of the wire/spring has a direct effect on the frequency. As the length increases, the frequency decreases and vice versa. This relationship is described by the formula:
f = 1 / (2π) * √(k/m) * (1/L)

Where L is the length of the wire/spring.

## 3. What is the significance of the spring constant in determining the frequency?

The spring constant is a measure of the stiffness of the spring. It determines how much force is required to stretch or compress the spring. A higher spring constant results in a higher frequency, while a lower spring constant results in a lower frequency.

## 4. How does the mass of the object attached to the spring affect the frequency?

The mass of the object attached to the spring also affects the frequency. A heavier mass will result in a lower frequency, while a lighter mass will result in a higher frequency. This relationship is described by the formula:
f = 1 / (2π) * √(k/m)

## 5. Can the frequency of a wire/spring system be changed?

Yes, the frequency of a wire/spring system can be changed by altering the length, spring constant, or mass of the object attached to the spring. Additionally, the frequency can also be changed by introducing a damping force, which reduces the amplitude of the oscillations and thus changes the frequency.

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