Frequency difference in two springs with different stiffness factors

[mhmil]

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.

 

[Doc Al]

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.

 

[tomcue]

I would suggest that your niece should first understand the relationship between stiffness and frequency in a spring-mass system. The frequency of a spring-mass system is directly proportional to the square root of the stiffness constant and inversely proportional to the square root of the mass. This means that a stiffer spring will have a higher frequency and a lighter mass will also result in a higher frequency.

To answer the question, we need to know the frequency of the prototype version. Without this information, it is not possible to determine the exact frequency of the new version. However, we can still compare the two versions in terms of their stiffness factors.

If the stiffness factor of the new spring is 2.2 times larger than the prototype spring, we can say that the new spring is 2.2 times stiffer. This means that the frequency of the new version will be higher than the prototype version, but we cannot determine the exact frequency without knowing the values of K and M.

In order to accurately calculate the frequency and compare the two versions, your niece will need to measure the stiffness constant and mass of both versions and plug them into the equation F= 1/2pi * square root of K/M. This will give her the actual frequency for each version and she can then compare them to see the difference.

In conclusion, a stiffer spring will have a higher frequency, but without knowing the values of K and M, we cannot accurately determine the exact frequency difference between the two versions. Encourage your niece to gather more data and make precise measurements in order to fully understand the relationship between stiffness and frequency in a spring-mass system.