What is the spring constant of a 129N object the vibrates at a period of 4.98s?
To find the spring constant of an object vibrating at a given period, we need to use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement of the spring.
Hooke's Law equation is expressed as:
F = -kx
Where:
F is the force applied on the spring (in newtons, N),
k is the spring constant (in newtons per meter, N/m),
and x is the displacement of the spring from its equilibrium position (in meters, m).
In this case, the object has a weight of 129 N, so we know that the force acting on it due to gravity is 129 N. When the object vibrates, the force exerted by the spring will be the same as the weight, but in the opposite direction.
So, we can rewrite Hooke's Law equation as:
129 N = -kx
Now, let's consider the period of oscillation, T. The period is the time it takes for one complete cycle of the vibration. In this case, the period is given as 4.98 seconds.
The period, T, is related to the angular frequency, ω, through the formula:
T = 2π/ω
We can rearrange this equation to solve for ω:
ω = 2π/T
Substituting the given period value, we find:
ω = 2π/4.98 s
Now, we know that the angular frequency, ω, is related to the spring constant, k, as follows:
ω = √(k/m)
Where m is the mass of the object attached to the spring. To find the mass, we can divide the weight by the acceleration due to gravity (g), which is approximately 9.8 m/s²:
m = 129 N / (9.8 m/s²)
Now, substituting the given values into the equation ω = √(k/m), we have:
2π/4.98 s = √(k / (129 N / (9.8 m/s²)))
To solve for the spring constant, k, we can square both sides of the equation:
(2π/4.98 s)² = k / (129 N / (9.8 m/s²))
Simplifying the right side, we get:
(2π/4.98 s)² = k / (129 N / 9.8 m/s²)
k = (2π/4.98 s)² * (129 N / 9.8 m/s²)
Calculating this expression will give you the value of the spring constant, k.