Why Does a Negative Spring Constant Give a Nonreal Period in Oscillation?

  • Thread starter Thread starter bmx_Freestyle
  • Start date Start date
  • Tags Tags
    Oscillation Period
AI Thread Summary
A negative spring constant indicates an error in the experimental setup or data collection, as it suggests an unstable system that cannot oscillate. The formula for the period of oscillation, T=2pi(m/k)^1/2, requires a positive k for real results. The discussion raises questions about the instruments used for measuring force and distance, as well as the type of experiment conducted, such as a pendulum or mass-spring system. Clarifying these aspects is essential to identify the source of the negative slope. Understanding the relationship between the variables and ensuring accurate measurements are crucial for obtaining valid results.
bmx_Freestyle
Messages
11
Reaction score
0
Ok so i am doing a lab...and i graphed my data points and my slope turned out to be negative.
The y-axis is in Newtons and the x-axis is in Meters.
So my teacher said the the slope=k (constant)
period of an oscillation formula: T=2pi(m/k)^1/2
But my k is negative so when i substitute all my numbers, i get a "Nonreal answer"
What did i do wrong?
 
Last edited:
Physics news on Phys.org
bmx_Freestyle said:
Ok so i am doing a lab...and i graphed my data points and my slope turned out to be negative.
The y-axis is in Newtons and the x-axis is in Meters.
So my teacher said the the slope=k (constant)
period of an oscillation formula: T=2pi(m/k)^1/2
But my k is negative so when i substitute all my numbers, i get a "Nonreal answer"
What did i do wrong?

You said this was a prac, and you were plotting the results.

What instrument were you using to measure the [Force, in] Newtons;

What instrument were you using to measure the metres;

that you eventually plotted?

What experiment were you actually doing? A pendulum? A mass on a spring? A mass in an inertial balance?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top