The body of 1275kg car is supported on a frame by four springs, two people riding in the car have a combined mass of 153kg when driven over a pathole in the road ,the frame vibrate with a period of 0.84sec for the first few seconds, the vibration approximates
Find K (spring Constant) of a single spring
To find the spring constant (K) of a single spring in this system, we need to use the formula for the period of oscillation (T) of a mass-spring system:
T = 2π√(m/k)
where:
T = period of oscillation
m = mass of the system
k = spring constant
In this case, the total mass of the car and the two people is given as 1275 kg + 153 kg = 1428 kg.
Since the system vibrates with a period of 0.84 seconds, we can plug in the values into the formula and solve for k:
0.84 = 2π√(1428/k)
To isolate k, we can square both sides of the equation and divide both sides by (2π)^2:
(0.84)^2 = (2π)^2 * (1428/k)
0.84^2 = 4π^2 * (1428/k)
Taking the square root of both sides:
√(0.84^2) = √(4π^2 * (1428/k))
0.84 = 2π * √(1428/k)
Now, we can isolate k by dividing both sides of the equation by (2π) and then square both sides:
(0.84 / (2π))^2 = (2π * √(1428/k)) / (2π)
(0.84 / (2π))^2 = √(1428/k)
Solving for the square root of the left side:
√[(0.84 / (2π))^2] = √(1428/k)
0.84 / (2π) = √(1428/k)
Squaring both sides, we get:
(0.84 / (2π))^2 = (1428/k)
Rearranging the equation, we can solve for k:
k = 1428 / (0.84 / (2π))^2
Calculating the right side of the equation:
k = 1428 / (0.84 / (2π))^2
k ≈ 226.3 N/m
Therefore, the spring constant (K) of a single spring in the system is approximately 226.3 N/m.