the weight of 50N suspend from a spring vibrate vertically with amplitude of 8cm angular frequency(natural frequency) of 1 oscillation\sec

Question

1)stiffness of the spring 2)find maximum velocity 3)find the maximum acceleration

fn=1\2pi root of k\m fn= natural frequency

w=mg

g=9.81

Answer 1

(1)

 

ω = √(k/m) ⇒ k = mω²

 

m = W/g = 50/9.8 kg ≅ 5.1 kg

 

ω = 1 oscillation/sec = 2π radians/sec

 

k = (5.1)(2π)² N/m ≅ 201.3 N/m

 

(2)

 

(1/2)mv_max² = (1/2)kA²

 

v_max = √(kA²/m) = A√(k/m) = Aω = (0.08)(2π) m/s = 0.503 m/s

 

(3)

 

a_max = F_max/m = kA/m = (201.3)(0.08)/5.1 m/s² ≅ 3.16 m/s²