How Does Cutting a Spring Affect Its Angular Frequency?

AI Thread Summary
Cutting a spring into two shorter sections results in each half having a spring constant that is double the original spring constant. For a 60-coil spring with a spring constant of 434 N/m, each 30-coil section will have a spring constant of 868 N/m. The angular frequency of the oscillation can be calculated using the formula ω = √(k/m), where k is the spring constant of the shorter spring and m is the mass attached. In this case, with a mass of 45.1 kg, the angular frequency can be determined using the adjusted spring constant. Understanding how cutting the spring affects its properties is crucial for solving the problem accurately.
JSmith89
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Homework Statement


A 60-coil spring has a spring constant of 434 N/m. It is cut into two shorter springs, each of which has 30 coils. One end of a 30-coil spring is attached to a wall. An object of mass 45.1 kg is attached to the other end of the spring, and the system is set into horizontal oscillation. What is the angular frequency of the motion?


Homework Equations


\omega=\sqrt{}k/m

if that didn't come out right, w(omega)=the square root of k/m

The Attempt at a Solution


I believe this is the right equation, but I don't know how to factor in the spring that is cut into two separate ones...like how/if it affects the spring constant or another variable. Thanks in advance! :-)
 
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spring!

JSmith89 said:
I believe this is the right equation, but I don't know how to factor in the spring that is cut into two separate ones...like how/if it affects the spring constant or another variable.

Hi JSmith89! :smile:

Look at it from first principles …

How is the spring constant defined?

Imagine the spring being compressed while still in one piece …

how is each half affected?

And so what is the spring constant of each half? :wink:
 
ok got it. each half will have a spring constant of 2k. thanks for your help!
 
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