Frequency difference in two springs with different stiffness factors

AI Thread Summary
The discussion revolves around calculating the frequency difference between two springs with varying stiffness constants. The original spring has a stiffness constant K, while the new spring has a stiffness constant of 2.2K. The frequency of oscillation is determined using the formula F = (1/2π)√(K/M). To find the ratio of the frequencies, one can set up a comparison using the stiffness constants, leading to the conclusion that the new frequency is √2.2 times the old frequency. This approach allows for understanding how changes in spring stiffness affect oscillation frequency without needing specific mass values.
mhmil
Messages
8
Reaction score
0

Homework Statement



Your niece is working on a unique science fair project - a clock built around a small mass vibrating on a little spring. Her prototype version is working ok, but it's running too slow. Given your experience in this course, you suggest to her that a stiffer spring should oscillate faster, so she goes out and finds a new one with stiffness constant ("K") that is a factor 2.2 times larger than the prototype spring. How many times larger is the frequency of the new version than the old one? (If you think the new frequency is 2 times the old one, enter 2.0)


Homework Equations



This is where the problem is lying for me. I'm attempting to use F = 1/2pi*square root of K/M. I'm unable to find the numbers to complete the K/M at all in this question.

The Attempt at a Solution



F= (1/2pi)=(.159)
(.159)*Square root of K/M

I cannot figure out what K and M should be and am clueless as to how I find the frequency out of this question and compare the two stiffness.
 
Physics news on Phys.org
You don't need the actual numbers. Set up a ratio of the frequencies. Hint: If the original spring constant is K, the new one is 2.2K.
 
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