A 1.10 N bird descends onto a branch that bends and goes into SHM with a period of 0.60 s. Determine the effective elastic force constant of the branch.
To determine the effective elastic force constant of the branch, we can use Hooke's Law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
First, let's recall the formula for the period of a mass-spring system in Simple Harmonic Motion (SHM):
T = 2π√(m/k)
where T is the period, m is the mass, and k is the force constant (also known as the spring constant).
We know the period (T) is 0.60 s, and we are given the weight (force due to gravity) of the bird, which is 1.10 N.
Since force (F) is equal to mass (m) multiplied by acceleration due to gravity (g), we can use the equation F = mg to determine the mass:
1.10 N = m * 9.8 m/s^2
Solving for m:
m = 1.10 N / 9.8 m/s^2
m ≈ 0.1122 kg
Now, we can rearrange the formula for the period to solve for the force constant (k):
T = 2π√(m/k)
Square both sides of the equation:
T^2 = 4π^2(m/k)
Rearrange and solve for k:
k = 4π^2m / T^2
Substitute the known values:
k = 4 * π^2 * 0.1122 kg / (0.60 s)^2
k ≈ 73.67 N/m
Therefore, the effective elastic force constant of the branch is approximately 73.67 N/m.