When a 85.0 kg. person climbs into an 1800 kg. car, the car's springs compress vertically 1.8 cm. What will be the frequency of vibration when the car hits a bump? (Ignore damping.)
To find the frequency of vibration when the car hits a bump, we need to use the equation for the frequency of a mass-spring system. The equation is:
f = (1 / 2π) * √(k / m)
where:
f is the frequency of vibration,
k is the spring constant,
and m is the mass of the system.
First, let's calculate the spring constant (k) using the given information that the car's springs compress vertically by 1.8 cm.
We can use Hooke's Law to find the spring constant, which states that the force exerted by a spring is directly proportional to the displacement, given by the equation:
F = -k * x
where:
F is the force,
k is the spring constant,
and x is the displacement.
In this case, the force applied by the springs equals the weight of the person climbing into the car, so we have:
F = m * g
where:
m is the mass of the person,
g is the acceleration due to gravity.
Given that the mass of the person is 85.0 kg, and the acceleration due to gravity is approximately 9.81 m/s², we can calculate the force (F):
F = m * g = 85.0 kg * 9.81 m/s²
Next, we can solve Hooke's Law equation for the spring constant (k):
F = -k * x
k = -F / x
Substituting the values we have:
k = -[(85.0 kg * 9.81 m/s²) / (0.018 m)]
Now that we have the spring constant, we can plug it into the equation for the frequency of vibration:
f = (1 / 2π) * √(k / m)
Substituting the values we have:
f = (1 / 2π) * √(k / 1800 kg)
Now we can calculate the frequency (f).