Tag Info

New answers tagged

2

Let's take $k=1$ for simplicity. Then our function $f(t)$ as a solution to equation $$\ddot f=-f-c \dot f,$$ $$f(0)=1,$$ $$\dot f(0)=0$$ will look like: $$f(t)=e^{-\frac{ct}2}\left(\cos\left(\frac12\sqrt{4-c^2}t\right)+\frac{c\sin\left(\frac12\sqrt{4-c^2}t\right)}{\sqrt{4-c^2}}\right).$$ Critically damped version of it is when we take limit $c\to2$: ...


0

Initial conditions do not matter with simple harmonic motion. The oscillating frequency is going to be found from the combined mass $\omega = \sqrt{\frac{k}{M+m}}$ To get the amplitude you take the general solution of $y(t) = y_0 + Y \sin \omega t$ and use it in the equations of motion $m \ddot{y} = (M+m) g- k\, y$ and solve for $Y$ and $y_0$ for all times ...


0

We use the coordinates $x_1$ and $x_2$ for the two blocks. Let $\ell$ be the natural length of the spring. Let us now write down the potential for this system. Note that if $x_2 - x_1 > \ell$ then the spring does not exert any force on either block and hence there is no potential. If $x_2 - x_1 < \ell$, then the spring is compressed with $\Delta x = ...


0

I think this test conditions are more about comparability than completely testing any possible road condition. When you test products, such as car suspensions, you design your test to be standardized. A standardized test gives you the possibility to compare results over various models and over time. If every suspension manufacturer would test its ...



Top 50 recent answers are included