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Smileyxx
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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 said: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?
Its a = \left( 2 \pi f\right)^{2} x.SHISHKABOB said:[itex]a = \left( 2 \pi f\right)^{2x}[/itex] ?
I don't recognize that equation >.> I'm very sorry
do you recognize
[itex]x \left( t \right) = Acos \left( \omega t - \phi \right)[/itex]
where
[itex]\omega = \sqrt{ \frac{k}{m}}[/itex]
is the angular frequency of the oscillation, A is the amplitude and [itex]\phi[/itex] is the phase shift?
Smileyxx said: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 said:[itex]a = \left( 2 \pi f\right)^{2}x[/itex] makes much more sense :)
angular frequency is just the natural frequency times 2π
or
[itex] \omega = 2 \pi f[/itex]
Stiffness is a measure of the resistance of a spring to deformation when a force is applied. It is a characteristic property of the spring and is determined by its material and physical dimensions.
Stiffness can be calculated by dividing the force applied to the spring by the displacement it experiences. It is represented by the letter "k" and has units of Newtons per meter (N/m) in the metric system.
Stiffness determines how much a spring will stretch or compress under a given force. A higher stiffness means that the spring will resist deformation more, resulting in a smaller displacement. This can affect the frequency and amplitude of oscillations in the system.
The stiffness of a mass spring system can be affected by the material properties of the spring, such as its modulus of elasticity, as well as its physical dimensions, such as its length and cross-sectional area. The number of springs in the system and their arrangement can also impact stiffness.
The stiffness of a mass spring system can be adjusted by changing the physical dimensions of the spring, such as its length or cross-sectional area. Additionally, different types of springs or materials with different stiffness values can be used in the system. The stiffness can also be altered by changing the number and arrangement of springs in the system.