Question

A body of mass m=1 that is hooked from an elastic spring with elastic constant k=2 satisfies the equation

mu′' +γu′ +ku=0
(a) How large should be the damping coefficient γ be such that the spring does not oscillate?
(b) How do you explain that a spring, which does oscillate in air, might not oscillate in water?

Answer 1

The damping coefficient should be equal to or greater than a certain value to prevent the spring from oscillating. The presence of damping forces in water can prevent a spring from oscillating.

Step-by-step explanation:

To ensure the spring does not oscillate, the damping coefficient γ should be equal to or greater than 2√(km), where k is the elastic constant and m is the mass. In this case, k=2 and m=1, so γ should be equal to or greater than 2√(2*1) = 2√2 = 2.83. Therefore, the damping coefficient γ should be at least 2.83 in order to prevent oscillation.

A spring may not oscillate in water due to the presence of damping forces caused by the viscosity of water. These damping forces decrease the amplitude of oscillation, eventually bringing it to a stop. In contrast, the air has less damping effect, allowing the spring to oscillate for longer periods of time.