A 2.20kg mass is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 21.3N is required to hold the mass at rest when it is pulled 0.190m from its equilibrium position (the origin of the x axis). The mass is now released from rest with an initial displacement of xi = 0.190m, and it subsequently undergoes simple harmonic oscillations. Calculate the force constant of the spring
k=F/x=21.3/0.19 = 112 N/m
To calculate the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
Hooke's Law can be written as:
F = -k*x
Where F is the force exerted by the spring, k is the force constant of the spring, and x is the displacement from the equilibrium position.
In this case, we know that when the mass is pulled 0.190m from its equilibrium position, a horizontal force of 21.3N is required to hold it at rest. Therefore, we can use this information to find the force constant.
Using Hooke's Law, we have:
21.3N = -k * 0.190m
Rearranging the equation to solve for the force constant:
k = -21.3N / 0.190m
k = -112.11 N/m
The force constant of the spring is approximately 112.11 N/m.