Question

A spring with a mass of 1 kg has a spring constant 43 kg/s2. If the spring begins at equilibrium position and is given a velocity of 5 m/s, find the damping constant that would produce critical damping.

Answer 1

Answer:

Damping constant is equal to $2\sqrt{43}$

Explanation:

We have given mass m = 1 kg

Spring constant is given $k=43kg/sec^2$

For critical damping $\zeta =1$

We have to find the damping constant

Damping constant is equal to $\zeta =\frac{c}{2\sqrt{\frac{k}{m}}}$

So $1 =\frac{c}{2\sqrt{\frac{k}{m}}}$

$c=2\sqrt{\frac{k}{m}}$

$=2\sqrt{\frac{43}{1}}=2\sqrt{43}$

So damping constant is equal to $2\sqrt{43}$