3 identical springs hanging from ceiling

  • Thread starter FlipStyle1308
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In summary, the three springs have the following constants: Spring 1 has a constant of k = 4pi^2(0.7143)/1, Spring 2 has a constant of k = 4pi^2(0.5)/1, and Spring 3 has a constant of k = -14N/m. The weight of the unknown object is 18.64 N.
  • #1
FlipStyle1308
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I got the following problem wrong, and don't know why, so hopefully somebody would be able to help me out!

Three identical (massless) springs are hung from the ceiling.
Spring 1: Nothing hung from spring. The spring's bottom is 0.2 m below the ceiling.
Spring 2: A 7 N object is hung from the spring. The spring's bottom is 0.5 m below the ceiling.
Spring 3: An object of unknown weight is hung from the spring. The spring's bottom is 1 m below the ceiling.

a. Find the spring constant that each spring has.
b. Find the weight of the unknown object.

For A, I used k = 4pi^2(0.7143)/1. I got 0.7143 as my mass by using F=ma, and I used 1 for T, since no time or frequency was given for either of the 3 springs. I guess since part A was wrong, that answer led to my answer for part B to be wrong also.
 
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  • #2
FlipStyle1308 said:
I got the following problem wrong, and don't know why, so hopefully somebody would be able to help me out!

Three identical (massless) springs are hung from the ceiling.
Spring 1: Nothing hung from spring. The spring's bottom is 0.2 m below the ceiling.
Spring 2: A 7 N object is hung from the spring. The spring's bottom is 0.5 m below the ceiling.
Spring 3: An object of unknown weight is hung from the spring. The spring's bottom is 1 m below the ceiling.

a. Find the spring constant that each spring has.
b. Find the weight of the unknown object.

For A, I used k = 4pi^2(0.7143)/1. I got 0.7143 as my mass by using F=ma, and I used 1 for T, since no time or frequency was given for either of the 3 springs. I guess since part A was wrong, that answer led to my answer for part B to be wrong also.
You are using the wrong equation. That equation can only be used when the mass is oscillating!

And I have no idea how you got your mass!

No, use that at equilibrium, k x = mg. You then find the spring constant from spring 2. Since they are identical, they all have the same k.
Now use that equation again to find the mass for spring 3 (the answer will be obvious)
 
  • #3
I got the same answer I had before. 14? Or should it be -14?
 
  • #4
FlipStyle1308 said:
I got the same answer I had before. 14? Or should it be -14?
14 N/m is the correct answer (k is never negative). Are you saying this is the wrong answer?

EDIT: Wait, I had not noticed the 0.2 meter of spring 1. So you must use a distance of 0.3 m, not 0.5m in the equation!
 
  • #5
Oh, so it's the distance below the original location of the spring without a mass?
 
  • #6
FlipStyle1308 said:
Oh, so it's the distance below the original location of the spring without a mass?
Yes. In the equation kx = mg, the x represents the *extra stretch* of the spring when the mass is attached as compared to when there was no mass at all.
 
  • #7
So 23.3 = k, and the weight of the unknown object is 18.64?
 
  • #8
Never mind, it's correct, thanks!
 

Related to 3 identical springs hanging from ceiling

1. How does the number of springs affect the overall system?

The number of springs affects the overall system by changing the stiffness and tension of the springs. With more springs, the system becomes stiffer and can support more weight. However, if the springs are not identical, the system may become unbalanced and cause uneven weight distribution.

2. What is the relationship between the length of the springs and their oscillation?

The length of the springs is directly related to their oscillation. The longer the springs, the slower their oscillation will be. This is because longer springs have a larger restoring force, making it harder for them to return to their equilibrium position.

3. Can the springs be replaced with different materials?

Yes, the springs can be replaced with different materials, but it may affect the overall system. Different materials have different stiffness and tension, which can alter the behavior of the system. It is important to use identical springs for accurate results.

4. What happens if one spring is removed or broken?

If one spring is removed or broken, the system will become unbalanced and the remaining springs will have to support more weight. This can cause the system to become unstable and may lead to unpredictable oscillations. It is important to always use all three identical springs for a stable system.

5. How can the oscillation of the system be changed?

The oscillation of the system can be changed by altering the length, stiffness, or tension of the springs. Shorter springs will have a faster oscillation, while stiffer and more tense springs will have a slower oscillation. Additionally, changing the weight or position of the object attached to the springs can also affect the oscillation of the system.

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