A 0.98 N bird descends onto a branch that bends and goes into SHM with a period of 0.46 s. Determine the effective elastic force constant of the branch.
To determine the effective elastic force constant of the branch, we need to use Hooke's Law. Hooke's Law states that the force exerted by an object (in this case, the branch) undergoing Simple Harmonic Motion (SHM) is directly proportional to the displacement from its equilibrium position.
The equation for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the branch (in newtons),
k is the effective elastic force constant of the branch (also known as the spring constant or stiffness) (in N/m),
x is the displacement from the equilibrium position (in meters),
and the negative sign indicates that the force is proportional to the opposite direction of the displacement.
In this problem, we're given the force (F) exerted by the bird as 0.98 N and the period (T) of the SHM as 0.46 s.
Let's start by finding the angular frequency (ω) of the SHM using the formula:
ω = 2π / T
where π is approximately equal to 3.14159.
ω = 2π / 0.46 s
ω ≈ 13.692 rad/s
The angular frequency (ω) can also be calculated using the formula:
ω = √(k / m)
where m is the mass of the bird.
Now, we need to find the mass of the bird using Newton's second law of motion:
F = m * g
Where g is the acceleration due to gravity (approximately 9.8 m/s²).
0.98 N = m * 9.8 m/s²
m ≈ 0.1 kg
Now we can use the angular frequency (ω) and the mass (m) to find the effective elastic force constant (k).
ω = √(k / m)
Squaring both sides:
ω² = k / m
Rearranging the equation:
k = ω² * m
Substituting the values:
k = (13.692 rad/s)² * 0.1 kg
k ≈ 18.856 N/m
Therefore, the effective elastic force constant of the branch is approximately 18.856 N/m.