0
$\begingroup$

Question:

enter image description here

Equations:

I'm having trouble understanding what a "solution" to equation 6 refers to? What are the implications of including gravity in the equation?

$\endgroup$
0
$\begingroup$

$Mg$ is a constant so only changes the equilibrium position, not the angular frequency. Equation 6 can be rearranged to give

$$\frac{d^2y}{dy^2}=-\frac{k}{M}(y+L_0+\frac{Mg}{k})$$

This makes the new equilibrium position $L_0+\frac{Mg}{k}$ .

$\endgroup$
0
$\begingroup$

It means does the solution of equation (5) satisfy eq (6)? You can check it by putting the solution in equation. As for your next question,it has just shifted the equilibrium point of the system around which your particle oscillated. You can gain more if you can solve the differential equation.

$\endgroup$
0
$\begingroup$

Mathematically this is a differential equation. But, a solution is any function that you write down and replace $y(t)$ with $guess(t)$ and $y(t)_{tt}$ with $guess(t)_{tt}$, and the equality will still hold.

The implication of including gravity is mathematically making it $F(y,y_{tt})=const \neq 0$ instead of having a homogeneous and boring $F(y,y_{tt})=0$. In this case you just need to add it to your guessed or calculated solution to the homogenous equation and it will be true. And it will also be a general solution, so cheating works.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.