Stiffness in mass spring system

[Smileyxx]

Homework Statement

What happens to the frequency of oscillation if stiffness increases and why?

Homework Equations

?

The Attempt at a Solution

Frequency increases but trying to figure out why.

 

[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?

 

[Smileyxx]

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

 

[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?

 

[Smileyxx]

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

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?

 

[SHISHKABOB]

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

$\omega = 2 \pi f$

 

[Smileyxx]

$a = \left( 2 \pi f\right)^{2}x$ makes much more sense :)

angular frequency is just the natural frequency times 2π

or

$\omega = 2 \pi f$

Thanks alot