12 points So far we have assumed that in a massspring system

[12 points] So far we have assumed that in a mass-spring system, the spring\'s mass is negligible compared to the mass attached to it. Let us consider what happens when the spring has a mass M that cannot be neglected. The spring’s equilibrium length is L0 and spring constant is k. When stretched or compressed to a length L, its potential energy is 1/2 k x2 where x = L - L0. a) Consider the massive spring by itself, with no additional mass attached. One end is fixed and the other end is moving horizontally with speed v. Assume that the speed of points along the length of the spring varies linearly with distance l from the fixed end. Assume also that the mass of the spring is distributed uniformly along the spring. Calculate the kinetic energy of the spring in terms of M and v. [Hint: divide the spring into equal-sized mass elements m within which the speed is approximately constant, figure out a formula for the speed vl of such a mass element located a distance l from the fixed end of the spring, and proceed]. b) For this part, let us return to the familiar example of a massless spring attached to a mass m. Taking the time derivative of the conservation of energy equation and using the fact that a = -2x, show that the angular frequency = (k/m). c) Going back to the massive spring of part a), and again assuming no additional mass m, use the same procedure as part b) to obtain an expression for the angular frequency of the massive spring. What is the effective mass of the spring?

Solution

Part A.) Let us consider a small mass dm at distance x from the origin and lenght dx. Now since the speed of the points varies linearly over the length such that the right end of the spring moves with velocity v, we can say that the particle at distance x would be moving at a speed of vx/L.
Also the mass of the particle of length dx would be Mdx/L

Therefore the kinetic energy of this small particle would be: dE = 0.5*(Mdx/L)*(vx/L)^2
or, dE = Mv²x² dx / L²

We can integrate the above expression over the length of the spring to get the total kinetic energy as:

E = Mv²x² dx / 2L³ = Mv²/6 is the required expression for total kinetic energy of the spring.

Part B.) Now for the massless spring, the energy equation is given as

U = 0.5 kx² + 0.5mv²

Now for conserved energy, we must have dU/dx = 0, differentiating the above expression, we getL:

kx = - mvdv/dx

Now, v can be written as dx/dt, hence we getL: kx = - m(dx/dt)(dv/dx) = - mdv/dt

Also, dv/dt = a = -²x, putting this in above equation we get:

kx = m²x

or = (k/m), which is the required expression.

Part C.) For the massive spring, again the net energy at any time would be:

U = 0.5kx² + mv²/6

Differentiating the above, we get:

dU/dx = 0 = kx + mvdv/dx/3

or, kx = - (m/3)dv/dt

or, kx = (m/3)²x

or, = (3k/m) is the required expression for the angular frequency, the effective mass is m/3