Why do I get a Warning: "cannot find explicit solution"?

I want to calculate in a mass-spring-damper system which "k0" value will be A<25. I have a following MatLab code:

clc
clear

syms k0
v0 = 2.5
c0 = 221229.947
m0 = 53.375
t = 0:0.01:1

alfa = sqrt(c0/m0)
beta = k0/(2*m0)
delta = sqrt(alfa^2 - beta^2)

A = v0/(sqrt(alfa^2 - (k0/(2*m0))^2)) .* exp(-(k0/(2*m0) .* t)) .* sin((sqrt(alfa^2 - (k0/(2*m0))^2)) .* t) * 10^3

solA = solve(A < 25, k0, 'MaxDegree', 4)

Thank you in advance.

Answer 1

[Walter Roberson]

t = 0:0.01:1

Your t is a vector.

A = v0/(sqrt(alfa^2 - (k0/(2*m0))^2)) .* exp(-(k0/(2*m0) .* t)) .* sin((sqrt(alfa^2 - (k0/(2*m0))^2)) .* t) * 10^3

Your A is in terms of t so your A is a vector.

solve(A < 25, k0, 'MaxDegree', 4)

That tries to solve for a single k0 that satisfies the entire vector of A simultaneously.

Also you need to know that when you solve() an inequality and do not use 'ReturnConditions', true then solve chooses a single representative value rather than describing the boundaries under which the inequality becomes true.

Also note that you did not confine to the real numbers so it will return complex solutions.

[Walter Roberson]

Your equation is non-linear. For particular k0 values k0 > 0, it has up to 21 different t solutions in the range 0 to 1. It is difficult to solve. It does not give you a quartic (degree 4 polynomial) for any particular t value because of the sin()

I do not think you will be able to proceed symbolically.

Answer 2

[Juhász Marcell]

How can I calculate "k0" value? I've also tried with two variables (k0, t) and I would seek the maximum from the derivative. But I didn't get a solution...