The springs of a 1100 kg car compress 6.0 mm when its 68 kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?
Can someone please explain to me how 2pi f = w = 10
f = 1.6 HZ
because this is the only part I can't figure out for myself.I highly appreciate it,thanks
6 mm = 6 * 10^-3 m
68 * 9.81 = 667 Newtons
k of spring = 667/(6*10^-3) = 111,167 N/m
w = sqrt(k/m) = sqrt (111,167/1100)
=10 radians/s
2 pi f = w = 10
f = 1.6 Hz
To calculate the frequency of vibrations when the car goes over a bump, we need to use Hooke's Law and the formula for the frequency of simple harmonic motion.
1. First, let's find the effective spring constant (k) of the car's springs.
Hooke's Law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. According to the problem, the springs compress by 6.0 mm (which we will convert to meters later) when the driver gets into the seat.
The formula for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.
Since we know the mass of the driver (68 kg) and the displacement (6.0 mm = 0.006 m), we can use the equation F = ma, where a is the acceleration due to gravity (9.8 m/s²) to find the force:
F = ma
F = (68 kg)(9.8 m/s²)
F = 666.4 N
Since F = kx, we can rearrange the equation to solve for k:
k = F/x
k = (666.4 N) / (0.006 m)
k ≈ 111,067 N/m
2. Next, let's calculate the natural frequency (f) of vibrations using the formula:
f = (1/2π) * √(k/m)
where f is the frequency, k is the spring constant, and m is the mass of the car (1100 kg).
Plugging in the values:
f = (1/2π) * √(111067 N/m / 1100 kg)
f ≈ 4.082 Hz
Therefore, when the car goes over a bump, the frequency of vibrations will be approximately 4.082 Hz.