Question

Consider the mass spring damper system my¨+ cy˙ + ky = u where y is the position of the mass m, the spring constant is k and c is the damping. Assume that all the initial conditions are zero, the input u(t) = 4 and the position y(t) = 1 − 2e−t + e−2t. Find m, c and k.

Answer 1

Step-by-step explanation:

Given a mass spring damper system

u = my'' + cy' + ky

where,

y is the position of the mass

m is the mass of the attached object

k is the spring constant

c is the damping.

All initial condition are zero.

Given that,

Input: u(t) = 4

Position: y(t) = 1 − 2e−t + e−2t

m, c and k?

Let find y' i.e. the dy/dt

Also, y'' i.e d²y/dt²

y(t) = 1 − 2e−t + e−2t

y'(t) = 0 + 2e-t — 2e-2t

y'(t) = 2e-t — 2e-2t

Also,

y"(t) = —2e-t + 4e-2t

Since

u(t) = my'' + cy' + ky

u(t) = m(-2e-t + 4e-2t) +c(2e-t - 2e-2t) + k(1 − 2e−t + e−2t)

Collect likes terms

u(t) = (4m - 2c + k)e-2t + (-2m + 2c -2k)e-t + k

Given that u(t) = 4

Therefore,

4 = (4m - 2c + k)e-2t + (-2m + c -2k)e-t + k

Comparing coefficient

The constant

k = 4 N/m

The e-2t is 0

4m - 2c + k =0

4m - 2c + 4 = 0

4m - 2c = -4

Divide both side by 2

2m - c = -2. Equation 1

Also the e-t is also 0

-2m+c-2k = 0

-2m + 2c -8 = 0

-2m + 2c = 8. Equation 2

Add equation 1 and 2 together

2m-c-2m+2c = -2+8

c = 6 kg/s

From equation 2

-2m+2c = 8

-2m = 8-2c

-2m = 8 - 2×6

-2m = 8 -12

-2m = -4

m = -4/-2

m = 2kg

Then, mass of the object attached is m = 2kg

The damping constant is c = 6 kg/s

The spring constant is k = 4N/m

Answer 2

The mass = m = 2 kg

The spring constant = k = 4 N/m

The damping = c = 6 kg/s

Step-by-step explanation:

u(t) = 4

y(t) = 1 - 2e⁻ᵗ + e⁻²ᵗ

my" + cy' + ky = u

y = 1 - 2e⁻ᵗ + e⁻²ᵗ

y' = 2e⁻ᵗ - 2e⁻²ᵗ

y" = -2e⁻ᵗ + 4e⁻²ᵗ

u = 4

Substituting all of these into the differential equation

my" + cy' + ky = u

m(-2e⁻ᵗ + 4e⁻²ᵗ) + c(2e⁻ᵗ - 2e⁻²ᵗ) + k(1 - 2e⁻ᵗ + e⁻²ᵗ) = 4

-2me⁻ᵗ + 4me⁻²ᵗ + 2ce⁻ᵗ - 2ce⁻²ᵗ + k - 2ke⁻ᵗ + ke⁻²ᵗ = 4

collecting like terms

(4m - 2c + k)e⁻²ᵗ + (-2m + 2c - 2k)e⁻ᵗ + k = 4

Comparing the RHS and the LHS

(4m - 2c + k)e⁻²ᵗ + (-2m + 2c - 2k)e⁻ᵗ + k = 0e⁻²ᵗ + 0e⁻ᵗ + 4

It is evident that;

4m - 2c + k = 0

-2m + 2c - 2k = 0

k = 4

Substituting the value of k into the first two expressions,

4m - 2c + k = 0

-2m + 2c - 2k = 0

4m - 2c + 4 = 0

-2m + 2c - 8 = 0

we obtain the simultaneous equation

4m - 2c = -4

-2m + 2c = 8

Solving the simultaneous equation,

m = 2 kg and c = 6 kg/s

Hence,

The mass = m = 2 kg

The spring constant = k = 4 N/m

The damping = c = 6 kg/s

Hope this Helps!!!