Question

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 4.42 Hz. (a) What is the spring constant of each spring if the mass of the car is 1450 kg and the mass is evenly distributed over the spring?

Answer 1

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².

Step-by-step explanation:

The period (T) of a spring in oscillation = 2π √(m/k)............. equation 1

Where m = mass acting on the spring (kg), k = spring constant of the spring (kg/s²).

Making k the subject of equation 1

k = T²/(4π²×m) .......................... equation 2

From the question, F = 4.42 Hz,

since T = 1/F

then, T = 1/F = 1/4.42 =0.226 s, π = 3.143

since the weight of the mass is evenly distributed over the four identical spring, Hence

m = 1450/4 = 362.5 kg

Substituting these values into equation 2

k = 0.226/{(4×3.143²)362.5}

k = 0.226/(14323.751)

k = 0.0000016 kg/s²

k = 1.6 × 10⁻⁶ kg/s².

Therefore, the spring constant of each spring = 1.6 × 10⁻⁶ kg/s².