- #1
Karol
- 1,380
- 22
Homework Statement
the two springs are identical and have the same constant k. they also have the same contracted length l0. a force of ##k\cdot l_0## has to be applied in order to start stretching the springs from length l0 and on.
L>l0, ##mg>k\cdot l_0##.
The ring slides frictionless on the vertical column.
Draw a graph of the force the spring applies as a function of it's elongation and express the force each spring applies as a function of it's length l.
Use the graph and write the expression for the elastic energy as a function of the spring's length l.
When the connection point between the springs A is at yA, what's the force the pipe exerts on the ring?
find y0, the location of the weight at equilibrium.
Express the potential energy of the 2 springs and weight system as a function of the mass's location y.
The weight is released from rest from point y=l0, where will it stop the first time?
What's it's maximum speed during it's movement?
Homework Equations
Elastic energy of spring: ##E=\frac{1}{2}kx^2##
Constant of an equivalent spring ##\over{k}## of two springs in a row: ##\frac{1}{\over{k}}=\frac{1}{k_1}+\frac{1}{k_2}##
The Attempt at a Solution
$$E=(l-l_0)kl_0+\frac{1}{2}k(l-l_0)^2=\frac{1}{2}k(l^2-l_0^2)$$
The force the horizontal spring applies: ##F_1=k\left( \sqrt{L^2+y_A^2} \right)##
The vertical one: ##F_2=k(y-y_A-l_0)##
$$F_{pipe}=F_1\cos\alpha=k\left( \sqrt{L^2+y_A^2} \right)\frac{L}{\left( \sqrt{L^2+y_A^2} \right)}=kL$$
The location of the mass at equilibrium y0 is found with 2 methods: the equivalent spring and a direct method.
The vertical force at yA the horizontal spring applies is:
$$F_{ver}=F_1\cos\alpha=k\left( \sqrt{L^2+y_A^2} \right)\frac{y_A}{\sqrt{L^2+y_A^2}}=ky_A$$
This is an equation of a linear spring, so the equivalent constant of the the "two" vertical springs is ##\frac{1}{2}k##
Using this spring y0 is:
$$mg=\frac{k}{2}y_0\;\rightarrow\; y_0=\frac{2mg}{k}$$
With the direct method is use the vertical force the horizontal spring exerts:
$$mg=F_1\sin\alpha\;\rightarrow\;mg=ky_A\;\rightarrow\;y_{0A}=\frac{mg}{k}$$
The other spring stretches also the same length so in total ##y_0=\frac{2mg}{k}##
The potential energy of the horizontal spring:
Since the vertical force of the horizontal spring spring at yA has the same constant k as the lower spring: ##y_A=\frac{y}{2}##
$$E_1=\frac{1}{2}k\left[ \left( \sqrt{L^2+y_A^2} \right)^2-l_o^2 \right]=\frac{k}{2}\left[ L^2+\frac{y^2}{4}-l_o^2 \right]$$
The potential energy of the vertical one:
$$E_1=\frac{1}{2}k\left[ (y-y_A)^2-l_0^2 \right]=\frac{k}{2}\left( \frac{y^2}{4}-l_0^2 \right)$$
I decide to make the gravitational energy of m relative to the equilibrium point ##\frac{2mg}{k}##:
$$E_p=mg\left[ y-\frac{2mg}{k} \right]$$
m is released from l0, where will it stop first? the potential energy at l0 is:
$$(E_1+E_2)_{l_0}=\frac{k}{2}\left[ L^2+\frac{l_0^2}{4}-l_0^2+\frac{l_0^2}{4}-l_0^2 \right]$$
$$(E_1+E_2)_{l_0}=E_p=\frac{k}{2}\left[ L^2-\frac{3}{2}l_0^2 \right]$$
This energy equals the gravitational energy:
$$E_p=mg\cdot y\;\rightarrow\;y=\frac{k}{2mg}\left[ L^2-\frac{3}{2}l_0^2 \right]$$
The mass makes a harmonic motion: ##y=A\cos(\omega t+\theta)##
$$A=\frac{2mg}{k}-l_0,\;\omega=\sqrt{\frac{2m}{k}}$$
The velocity is max when m passes through the equilibrium point, where ##\sin=0##:
$$\dot x=-\omega A\sin(\omega t+\theta)\;\rightarrow\; \dot x=\sqrt{\frac{2m}{k}}\left( \frac{2mg}{k}-l_0 \right)$$