You do it by using what you are given. Unfortunately, you haven't mentioned
what that is, so we can't be any more specific.
F = - k x In this equation, x is the distance that the spring has been stretched or compressed away from its equilibrium position F is the restoring force exerted by the spring. k is the spring constant.
angular frequency = square root (K/m) wher k is spring constant and m = mass linear frequency = 1/2pi times square root (K/m)
It is Newtons per metre.
k is often a constant. We don't know what the constant is, but we know we could find out.
Rate of flow varies as R^4 where R is the radius or Rate of flow = (k) x (R^4)
Measure the force (f) required to compress the spring a given amount (x) then use hooke's law to compute the spring constant (k) (f=kx)
F = - k x In this equation, x is the distance that the spring has been stretched or compressed away from its equilibrium position F is the restoring force exerted by the spring. k is the spring constant.
The spring constant is a measure of stiffness - the ability to resist displacement under a load. It is denoted by K where F = kx where f = load force and x = displacement
2k
You can find out how long a spring has been stretched/compressed by knowing it's elastic constant and the force the spring is exerting trying to go back to it's original shape. F=K*x (Moore's law) F is the force exerted by the spring. K is the elastic constant. X is the displacement of the end of the spring from it's normal position. You want to find x, x = F/K
Hooke's law was designed to determine the restoring force of a spring, given its spring constant and the displacement of the spring from its equilibrium position. The law is written as follows: F = -kx; in which "F" is the restoring force, "k" is the spring constant, and "x" is the spring's displacement.
It takes a larger force to compress or pull a spring the same distance as a spring with a smaller spring constant. This is shown in Hooke's law. x=F/k k---is the spring constant F---is the force applied to the spring x is the distance the spring has been compressed
Hooks law: F=-x*K F=Force x=distance = 0.5m K=constant F=mass*9.81 = 30*9.81=aaaa [N] aaa = -0.5 * K => K = something
the relation between force (F) and elongation (x) is F = kx where k is the spring constant. The stiffer the spring, the higher the force needed to get a certain elongation; or, for a given force, the elongation will be less for a stiffer spring
angular frequency = square root (K/m) wher k is spring constant and m = mass linear frequency = 1/2pi times square root (K/m)
It is Newtons per metre.
x represents displacement or deflection or distance. Hooke law states that force = k times x; the higher the displacement, x, for a given spring constant (k).