When a 69 kg person climbs into a 1000 kg car, the car's springs compress vertically by 2.9 cm. What will be the frequency of vibration when the car hits a bump? Ignore damping.
answer should be given in Hz.
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To calculate the frequency of vibration when the car hits a bump, we can use the formula for the frequency of a mass-spring system:
f = (1 / (2π)) * √(k / m)
Where:
f = frequency of vibration
π = 3.14159 (the mathematical constant pi)
k = spring constant
m = mass of the system
Now let's break down the problem to find the required values:
1. Calculate the spring constant (k):
The spring constant can be determined using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
F = kx
Where:
F = force exerted by the spring
k = spring constant
x = displacement from the equilibrium position (compression)
Given that the car's springs compress vertically by 2.9 cm (or 0.029 m), we can substitute the values into the equation:
F = k * 0.029
The force exerted by the spring is equal to the weight of the person, which is the mass multiplied by acceleration due to gravity (g = 9.8 m/s^2):
F = mg
F = 69 kg * 9.8 m/s^2
Setting the two equations equal to each other, we have:
mg = k * 0.029
Solving for k:
k = (mg) / 0.029
2. Calculate the frequency (f):
Now that we have the spring constant (k) and the mass (m) of the system (car + person), we can substitute these values into the formula for the frequency:
f = (1 / (2π)) * √(k / m)
Substituting the calculated values, we get:
f = (1 / (2π)) * √((mg) / (0.029 * m))
Simplifying the equation:
f = (1 / (2π)) * √((g) / (0.029))
Now, we can plug the value of g (9.8 m/s^2) into the equation and calculate the frequency of vibration (f).