A car has a mass of 1405 kg. hits a bump, and oscillates 0.669
times per second. What is the
spring constant of the shock
absorbers?

24958.764 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1405 kg

Frequency (f) = 0.669 s¯¹

Spring constant (K) =?

Next, we shall determine the period. This can be obtained as follow:

Frequency (f) = 0.669 s¯¹

Period (T) =?

T = 1/f

T = 1/0.669

T = 1.49 s

Finally, we shall determine the spring constant. This can be obtained as follow:

Mass (m) = 1405 kg

Period (T) = 1.49 s

Pi (π) = 3.14

Spring constant (K) =?

T = 2π√(m/K)

1.49 = 2 × 3.14 × √(1405/K)

1.49 = 6.28 × √(1405/K)

Divide both side by 6.28

1.49 / 6.28 = √(1405/K)

Take the square of both side

(1.49 / 6.28)² = 1405/K

2.2201 / 39.4384 = 1405/K

Cross multiply

2.2201 × K = 39.4384 × 1405

Divide both side by 2.2201

K = 39.4384 × 1405 / 2.2201

K = 24958.764 N/m

Thus, the spring constant is 24958.764 N/m

no. dot product   =0 when vecs at right angles