6.11Exercises, examples & solutions

Sketch; $z=0$ is when the mass is $l_0$ below the ceiling.

Equilibrium position of the mass $m$:

Force of the spring: $F_v = -k(l-l_0) = -kz$. Thus

Equation of motion for $m$: set up N2

Solution with $z(0) = z_{eq}$ and $v(0) = v_0$:

homogeneous part of the equation: $m\frac{dv}{dt} + kz=0$

with $\omega_0^2 = \frac{k}{m}$

special solution: $z_s = - \frac{mg}{k}$

general solution:

initial conditions:

and

Thus, the solution is

Sketch; $z=0$ is when the mass is at $l_0$ below the ceiling. Now we have 2 springs, one with spring constant $k_1$, the other with $k_2$. Both have the same rest length $l_0$

Equilibrium position of the mass $m$:

Forces of the springs: $F_{v1} = -k_1(l-l_0) = -k_1 z$ and $F_{v2} = -k_2(l-l_0) = -k_2 z$. Thus

Equation of motion for $m$: set up N2

Thus we conclude, that the exercise is basically the same: all we have to do is replace $k$ by $K_{tot} = k_1 + k_2$

The solution with $z(0) = z_{eq}$ and $v(0) = v_0$ is thus

with $\omega_0^2 = \frac{k_{tot}}{m}$

Again, we have two springs acting on the mass. However, they are no on opposite sides. We expect on symmetry arguments that the equilibrium will be in the middle, i.e at $x=0$.

If the mass is positioned to the right of $x=0$, spring 1 is extended beyond its rest length and will pull in the negative $x$-direction:

Spring 2 will than be shorter than its rest length and will push to the negative $x$-direction:

Thus, equilibrium is reached when

as we anticipated.

Equation of motion for $m$: set up N2

Thus we conclude, that the exercise is basically the same: all we have to do is replace $k$ by $k_{tot} = 2k$

General solution $x(t) = A \sin \omega_0 t + B \cos \omega_0 t$ with $\omega_0^2 = \frac{2k}{m}$.

Like in the previous exercises, it is now a matter of specifying the initial conditions and finding $A$ and $B$.

Again, we have two springs acting on the mass. Now they don’t fit both with their rest length. The will be compressed and try to lengthen. However, based on symmetry we still do expect that $x=0$ is the equilibrium position.

If the mass is positioned to the right of $x=0$, spring 1 stille too short and will push to the right:

Spring 2 will than be even shorter and will push to the negative $x$-direction:

Thus, equilibrium is reached when

as we anticipated.

Equation of motion for $m$: set up N2

Thus we conclude, $k_{tot} = 2k$, which is identical to the previous exercise!