Question

A mass suspended at the end of a light Spring and spring constant K is set into vertical motion use lagrange's equation to find the equation of motion of the mass

Answer 1

the general form of the Lagrangian:

L = T ? V

T is the kinetic energy of the system. The kinetic energy is equal to 1

2

mv

2 where m is the mass, and v is

the velocity of the mass. There is no surprise yet, and it comes as no surprise either that =

(vx

2 + vy

2 + vz

2

)

1

2

. So letting vx = x? and so on, were the dot represents the time derivative we have the

kinetic energy of the system T =

1

2

m(x?

2 + y?

2 + z?

2

) and we are now half way done.

The potential energy of the system consists of two parts: The elastic potential energy stored in the

displacement of the spring from its equilibrium position, and the gravitational potential energy. Since the

Lagrangian is derived for the express purpose of differentiating it we will not be concerned with the details

of defining the gravitational potential energy, namely the discrepancies regarding the reference for origin

and how that effects the values of the potential energy.

Starting with the elastic potential energy, we have the pleasure of the

simplicity of Hooke’s Law F = ?kx which we can integrate over a

displacement to get Velastic =

1

2

kx

2

. To make this apply to our model, all we

need to do is imply that there is a length that the spring with the mass

attached to it, will rest at equilibrium stretched under the weight of the

mass, mg, we will call this length l. Furthermore we will consider a length l0

that is the natural, unstretched length of the spring with no weight attached. By Hooke’s law the

relationship between these lengths is k(l ? l0

) = mg. Since l0 is a constant characteristic of the system

considered, we conclude that Velastic = Ve =

1

2

k(r ? l0

)

2 where r = (x

2 + y

2 + z

2

)

1

2.

Lastly, with all of the simplicity expected the gravitational potential energy of the system is Vg = mgz.

Things brings us to a complete Lagrangian for the system with the assumptions proposed in section 1:

L =

1

2

m(x?

2 + y

2 + z?

? 2

)

1

2 ? [

1

2

k(r ? l0

)

2 + mgz]

By differentiating with respect to time with the chain rule, and equating functions of each variable to their

time derivatives we get a system of ordinary differential equations:

x?= ??z

2

r ? l0

r

x

y?= ??z

2

r ? l0

r

y

z?= ??z

2

r ? l0

r

z ? g

Where ?z

2 =

k

m

is referred to with subscript of z on the character ? to indicate that it does not represent

a frequency of the system, but only the frequency of the spring-mass system with only vertical oscillations.

The system derived above has three degrees of freedom in space, but only two constants of motion, the

total energy, E = T + V, in the system and the angular momentum per unit mass about the vertical

represented: h = xy? ? yx?; this accounts for two invariant forms of the system. Therefore we say that the

system is not integrable in general, though in some regimes, particularly those of small oscillations,

“approximate analytical solutions can be found”