An 90 kg person steps into a car of mass 2600 kg, causing it to sink 2.35 cm on its springs. Assuming no damping, with what frequency will the car and passenger vibrate on the springs?
To find the frequency at which the car and passenger vibrate on the springs, we can use Hooke's Law and the equation for the frequency of a mass-spring system.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement. Mathematically, it can be represented as:
F = -kx
Here, F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
We know that the car sinks 2.35 cm (or 0.0235 m) on its springs. The force exerted by the springs can be calculated using the mass and acceleration due to gravity:
F = mg = (90 kg)(9.8 m/s^2) = 882 N
According to Hooke's Law, the force exerted by the springs is equal to the force applied on them by the person:
F = -kx
882 N = -k(0.0235 m)
Solving for the spring constant (k):
k = -882 N / 0.0235 m = -37531.91 N/m
Now, we can use the equation for the frequency of a mass-spring system:
f = (1 / 2π) * sqrt(k / m)
where f is the frequency, k is the spring constant, and m is the mass.
Plugging in the values:
f = (1 / 2π) * sqrt(-37531.91 N/m / 2600 kg)
f ≈ 1.032 Hz
Therefore, the car and passenger will vibrate on the springs at a frequency of approximately 1.032 Hz.
To find the frequency at which the car and the passenger will vibrate on the springs, we need to use Hooke's law and the equation for the frequency of a simple harmonic oscillator.
Hooke's law states that the force exerted by a spring is directly proportional to its displacement. Mathematically, it can be expressed as:
F = -kx
where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the displacement is given as 2.35 cm (which is 0.0235 m), and the mass of the person is 90 kg. We need to find the spring constant, k.
The force exerted by the person on the car can be calculated using the person's weight, considering that acceleration due to gravity is 9.8 m/s²:
Force = mass * acceleration due to gravity
Force = 90 kg * 9.8 m/s²
Force = 882 N
Since the spring force is equal to the force exerted by the person on the car, we can equate the two:
-kx = 882
Rearranging the equation, we find:
k = -882 / x
Now we have the spring constant, k. To find the frequency, we can use the equation for the frequency of a simple harmonic oscillator:
frequency = 1 / (2π √(m / k))
where m is the mass of the car (2600 kg).
Substituting the values, we get:
frequency = 1 / (2π √(2600 / (-882 / 0.0235)))
Calculating this expression will give us the frequency at which the car and passenger will vibrate on the springs.
The spring constant is
k = (90*9.8N/0.0235 m) = 3.75^10^4 N/m
and the vibration frequency is
w = [1/(2 pi)]*sqrt(k/Mtotal)
Mtotal = 2690 kg
w = ___ s^-1