Odd equation relating to springs and SHM

In summary, the conversation revolves around a lab report for 12th grade physics, specifically finding the value of the force constant of a spring. The lab involves measuring the number of oscillations a weight makes on the end of the spring and using the equation T = 2pi (m/k)^1/2 to find the spring constant. The teacher then asks the students to find the value of m/3 using a graph of T^2 vs m and the equation T = 2pi [(M + m/3)/k]^1/2. The expert summarizer explains that the only difference between the two equations is the value for mass and suggests using the plot from the original lab to calculate the new mass. This process is
  • #1
K - Prime
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Well I am working on a lab report for 12th grade physics and my teacher has decided to throw a cuve ball at us. The lab is designed to determine the value of the force constant of a spring by measuring the number of oscillations a certain weight makes when on the end of the spring. To find k was easy, i just used the equation T = 2pi (m/k)^1/2. Heres the part i don't get...he asked us to find the value of m/3 using a graph of T^2 vs m (mass on spring) and gave us the equation T = 2pi [(M + m/3)/k]^1/2. All i can say is WHAT!?
 
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  • #2
Well the point ofthe lab wasto find the spring constant, and you modeled that already by graphing it, correct? (If not, do so). Now since you know the trend of period vs mass, you should be able to predict mass ifyou measure a certain period. Notice that the only difference between the firstequation you gave and the last one (which you didnt recognize) is that the last one has a different value for mass.

[tex] T = 2\pi\sqrt{\frac{m}{k}} [/tex]

He gives:

[tex] T = 2\pi\sqrt{\frac{M+m/3}{k}} [/tex]

If you say the mass is [itex] n [/itex], then [itex] n = M + m/3 [/itex] and the second equation becmoes

[tex] T = 2\pi\sqrt{\frac{n}{k}}[/tex], which is analogous to the first equation. So the numerator under the square root is simply a new total mass, nothing to be surprised about. Using the plot you derived in your original lab, find a way to calculate the new mass.

This process isn't necessary to do the problem, but hopefully it helps you see what is going on more clearly.
 
  • #3
ahh i see, that helps alot! :smile: thx!
 

Related to Odd equation relating to springs and SHM

1. What is SHM and how does it relate to springs?

SHM stands for Simple Harmonic Motion, and it is a type of motion where an object oscillates back and forth around a central equilibrium point. In the case of springs, SHM occurs when the spring is stretched or compressed and then released, causing it to vibrate back and forth.

2. What is the equation for calculating the period of a spring in SHM?

The equation for calculating the period of a spring in SHM is T = 2π√(m/k), where T is the period (time for one complete oscillation), m is the mass of the object attached to the spring, and k is the spring constant (a measure of the stiffness of the spring).

3. What does the 'k' value represent in the equation for SHM?

The 'k' value represents the spring constant, which is a measure of how stiff the spring is. A higher k value indicates a stiffer spring, while a lower k value indicates a more flexible spring.

4. Can the equation for SHM be used for any type of spring?

Yes, the equation for SHM can be used for any type of spring, as long as the motion is simple harmonic. This means that the restoring force (force that brings the object back to equilibrium) is directly proportional to the displacement (distance from equilibrium) and acts in the opposite direction of the displacement.

5. How can the equation for SHM be applied in real-world situations?

The equation for SHM can be applied in many real-world situations, such as in the design of suspension systems for cars, the movement of pendulums, and the vibrations of musical instruments. It can also be used to study and understand natural phenomena, such as the motion of tides and the behavior of atoms in a solid material.

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