# Stiffness in mass spring system

by Smileyxx
Tags: mass, spring, stiffness
 P: 34 1. The problem statement, all variables and given/known data What happens to the frequency of oscillation if stiffness increases and why? 2. Relevant equations ? 3. The attempt at a solution Frequency increases but trying to figure out why.
 P: 615 By stiffness I'm assuming you mean the spring constant, often known as k. It's found in Hooke's law F = -kx where F is the force the spring is exerting and x is the distance from the equilibrium point. Don't you have some equations relating the frequency of the oscillations of the spring-mass system and the spring constant of the spring?
P: 34
 Quote by SHISHKABOB By stiffness I'm assuming you mean the spring constant, often known as k. It's found in Hooke's law F = -kx where F is the force the spring is exerting and x is the distance from the equilibrium point. Don't you have some equations relating the frequency of the oscillations of the spring-mass system and the spring constant of the spring?
a=[2pie f]^2x. So if k increases f increases which increases a(acceleration) and according to the formula given at the start frequency increases. If that make sense

P: 615

## Stiffness in mass spring system

$a = \left( 2 \pi f\right)^{2x}$ ?

I don't recognize that equation >.> I'm very sorry

do you recognize

$x \left( t \right) = Acos \left( \omega t - \phi \right)$

where

$\omega = \sqrt{ \frac{k}{m}}$

is the angular frequency of the oscillation, A is the amplitude and $\phi$ is the phase shift?
P: 34
 Quote by SHISHKABOB $a = \left( 2 \pi f\right)^{2x}$ ? I don't recognize that equation >.> I'm very sorry do you recognize $x \left( t \right) = Acos \left( \omega t - \phi \right)$ where $\omega = \sqrt{ \frac{k}{m}}$ is the angular frequency of the oscillation, A is the amplitude and $\phi$ is the phase shift?
Its a = \left( 2 \pi f\right)^{2} x.
I dint knew the 2nd equation but i get it now. stiffness is proportional to frequency according to that.but does the natural frequency of spring increases as well?
P: 615
 Quote by Smileyxx Its a = \left( 2 \pi f\right)^{2} x. I dint knew the 2nd equation but i get it now. stiffness is proportional to frequency according to that.but does the natural frequency of spring increases as well?
$a = \left( 2 \pi f\right)^{2}x$ makes much more sense :)

angular frequency is just the natural frequency times 2π

or

$\omega = 2 \pi f$
P: 34
 Quote by SHISHKABOB $a = \left( 2 \pi f\right)^{2}x$ makes much more sense :) angular frequency is just the natural frequency times 2π or $\omega = 2 \pi f$
Thanks alot

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