When a 81.0 kg. person climbs into an 1400 kg. car, the car's springs compress vertically 1.5 cm. What will be the frequency of vibration when the car hits a bump? (Ignore damping.)

f = (1/2pi)*sqrt(k/M) Hz

M = 1481 kg
k = 81*9.8/0.015 = 52900 N/m

f = 5.0 Hz.

To find the frequency of vibration when the car hits a bump, we need to apply the principles of simple harmonic motion.

First, let's calculate the effective spring constant of the car's springs. The compressed vertical displacement of the car's springs is 1.5 cm.

The formula for the effective spring constant is given by:
k_eff = k1 + k2

where k1 is the initial spring constant of the car's springs and k2 is the spring constant due to the person's weight.

Assuming the car's springs have a linear relationship, we can calculate k1 using Hooke's Law:

k1 = F1 / x

where F1 is the force applied on the car's springs and x is the displacement.

The force applied on the car's springs due to the car is given by:
F1 = m1 * g

where m1 is the mass of the car and g is the acceleration due to gravity.

Plugging in the values, we get:
F1 = (1400 kg) * (9.8 m/s^2) = 13720 N

Using the formula for the spring constant, we have:
k1 = (13720 N) / (0.015 m) = 914666.67 N/m

Now, let's calculate k2 using the same formula:
k2 = F2 / x

The force applied on the car's springs due to the person is given by:
F2 = m2 * g

where m2 is the mass of the person.

Plugging in the values, we get:
F2 = (81.0 kg) * (9.8 m/s^2) = 794.8 N

Using the formula for the spring constant, we have:
k2 = (794.8 N) / (0.015 m) = 52986.67 N/m

Now that we have the values of k1 and k2, we can calculate the effective spring constant:
k_eff = k1 + k2 = 914666.67 N/m + 52986.67 N/m = 967653.34 N/m

The formula for the frequency of vibration in simple harmonic motion is given by:
f = (1 / 2π) * √(k_eff / m)

where f is the frequency of vibration, k_eff is the effective spring constant, and m is the total mass of the system (car + person).

Plugging in the values, we get:
f = (1 / 2π) * √(967653.34 N/m / (1400 kg + 81.0 kg))
f = 3.31 Hz

Therefore, the frequency of vibration when the car hits a bump, ignoring damping, is approximately 3.31 Hz.