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Liam B

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Liam

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- Thread starter Liam B
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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.

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Liam B

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Liam

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- #2

DEvens

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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?

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Liam B

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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.

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:

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nasu

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And how did you measure the tension?

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Liam B

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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.nasu said:And how did you measure the tension?

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.

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.

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.

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)

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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