Differential equations
a 2 kg mass on a spring with spring constant k = 1 is not oscillating. the system is critically damped. what is the coefficient of friction?

Step-by-step explanation:

For the spring mass system

my"+by'+ky=0

The system is said to be critically damped if

Here b is coefficient of friction

m is the mass of system

And k is spring constant

We have given m = 2 kg , k =1

So

So the coefficient of friction is

c > -10.55

step-by-step explanation:

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!