A mass 'm' is attached to a spring with spring constant 'k' and experiences a damping force F d = − b v , where 'v' is the velocity and 'b' is the damping coe

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A mass 'm' is attached to a spring with spring constant 'k' and experiences a damping force F d ​ = − b v , where 'v' is the velocity and 'b' is the damping coefficient. For the system to be critically damped, the damping coefficient 'b' must be equal to:

TestHub · Accepted Answer

For a damped harmonic oscillator, the equation of motion is $m \frac{d ^{2} x}{d t ^{2}} + b \frac{d x}{d t} + k x = 0$ .

Comparing this to the standard form $m \frac{d ^{2} x}{d t ^{2}} + 2 λm \frac{d x}{d t} + k x = 0$ , we have $2 λm = b$ , so the damping coefficient per unit mass is $λ = \frac{b}{2 m}$ . The natural angular frequency of the undamped oscillator is $ω_{0} = \frac{k}{m}$ . For critical damping, the condition is $λ = ω_{0}$ . Substituting the expressions for $λ$ and $ω_{0}$ : $\frac{b}{2 m} = \frac{k}{m}$ . Solving for 'b': $b = 2 m \frac{k}{m} = 2 m^{2} \frac{k}{m} = 2 mk$ .

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