Answer:
a) 1.184 × 10⁴N/m
b) 0.888 Hz
Step-by-step explanation:
a) 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 2.00 Hz.
The angular frequency of the vertical oscillation of the car is
ω = 2πf = 4π rad/s
The spring constant of the car is
ω = √(ktot/M) ⇒ ktot = Mω²
= 4.935 × 10⁴ N/m
Four identical springs support the car. Therefore, each spring has a spring constant of
k = ktot/4 = 1.184 × 10⁴N/m
b) Using the mass of 76kg
The spring constant of the car does not change. However, the mass of the oscillator changes due to
the four passengers. The oscillation frequency will be
f = ω/(2π) = (1/(2π))√(ktot/(M + 4m)) = 0.888 Hz