A 85.0 kg person steps into a car of mass 2500.0 kg, causing it to sink 2.5 cm. Assuming no damping, with what frequency will the car and passenger vibrate on the springs?
Have you calculated the spring constant?
Period=1/frequency= ....
To find the frequency at which the car and passenger will vibrate on the springs, we need to use Hooke's Law and the equation for the natural frequency of a mass-spring system.
First, let's find the spring constant of the car's suspension system using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement:
F = -k * x
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position. Since the car sinks 2.5 cm (or 0.025 m) under the weight of the person, the force exerted by the spring is given by:
F = m * g
Where m is the mass of the person and g is the acceleration due to gravity. Plugging in the values, we have:
F = (85.0 kg + 2500.0 kg) * 9.8 m/s²
F = 2585.0 kg * 9.8 m/s²
F = 25333 N
Now we can solve for the spring constant k:
25333 N = -k * 0.025 m
k = -25333 N / 0.025 m
k = -1013320 N/m
Next, we can use the equation for the natural frequency of a mass-spring system:
ω = √(k / m)
Where ω is the angular frequency (2π times the frequency) and m is the mass of the car with the person. Plugging in the values, we have:
ω = √(-1013320 N/m / (85.0 kg + 2500.0 kg))
ω = √(-1013320 N/m / 2585.0 kg)
ω = √(-391.93 N/kg)
ω = 19.8 rad/s
Finally, we can convert the angular frequency to frequency:
f = ω / (2π)
f = 19.8 rad/s / (2π)
f ≈ 3.15 Hz
Therefore, the car and passenger will vibrate on the springs with a frequency of approximately 3.15 Hz.