Answer:
the frequency of the oscillation is 1.5 Hz
Explanation:
Given;
mass of the spring, m = 1500 kg
extention of the spring, x = 5 mm = 5 x 10⁻³ m
mass of the driver = 68 kg
The weight of the driver is calculated as;
F = mg
F = 68 x 9.8 = 666.4 N
The spring constant, k, is calculated as;
k = F/m
k = (666.4 N) / (5 x 10⁻³ m)
k = 133,280 N/m
The angular speed of the spring is calculated;

The frequency of the oscillation is calculated as;
ω = 2πf
f = ω / 2π
f = (9.426) / (2π)
f = 1.5 Hz
Therefore, the frequency of the oscillation is 1.5 Hz
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
Answer:
DO NO KNOW AND I HOPE YOU CAN FIND IT
GOOD LUCK
Explanation:
Answer:
350 N/m
Explanation:
If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m
k = 70 / 0.2 = 350 N/m
The spring constant is 350 N/m