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A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring’s mass, consider a spring with mass M, equilibrium length \({L_0}\), and spring constant k. When stretched or compressed to a length L, the potential energy is \(\frac{1}{2}k{x^2}\), where \(x = L - {L_0}\).

(a) Consider a spring, as described above that has one end fixed and the other end moving 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 Mof the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of M and v. (Hint: Divide the spring into pieces of length dl; find the speed of each piece in terms of l, v, and L; find the mass of each piece in terms of dl, M, and L; and integrate from 0 to L. The result is not \(\frac{1}{2}M{v^2}\), since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass m moving on the end of a massless spring. By comparing your results to Eq. (14.8), which defines \(\omega \), show that the angular frequency of oscillation is \(\omega = \sqrt {\frac{k}{m}} \). (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega \) of the spring considered in part (a). If the effective mass M′of the spring is defined by \(\omega = \sqrt {\frac{k}{{M'}}} \), what is M′in terms of M?

Step by step solution

01

Given Data

Mass of the spring = M

Equilibrium Length = \({L_0}\)

Spring Constant = k

The potential energy = \(\frac{1}{2}k{x^2}\) \(x = L - {L_0}\)

02

Concept

Force constant is defined as the spring’s resisting force directed towards the equilibrium.

03

Step 3(a): Calculate the kinetic energy of the spring

Mass of small part of the spring is,

\(dm = \frac{M}{L}dl\)

Velocity at any point on the spring is,

\(u = v\left( {\frac{l}{L}} \right)\)

Kinetic energy of small part of the spring is,

\(dK = \frac{1}{2}dm{u^2}\)

Substitute \(dm = \frac{M}{L}dl\;{\rm{and}}\;u = v\frac{l}{L}\)

\(\begin{array}{c}dK = \frac{1}{2}\left( {\frac{M}{L}dl} \right){\left( {\frac{{vl}}{L}} \right)^2}\\ = \frac{1}{2}\left( {\frac{{M{v^2}}}{{{L^3}}}} \right){l^2}\;dl\end{array}\)

Total kinetic energy,

\(\begin{array}{c}K = \int {dK} \\ = \frac{1}{2}\left( {\frac{{M{v^2}}}{{{L^3}}}} \right)\int {{l^2}\;dl} \\ = \frac{1}{2}\left( {\frac{{M{v^2}}}{{{L^3}}}} \right)\left( {\frac{{{L^3}}}{3}} \right)\\ = \frac{1}{2}\left( {\frac{{M{v^2}}}{3}} \right)\\K = \frac{{M{v^2}}}{6}\end{array}\)

Hence the kinetic energy of the spring is \(K = \frac{{M{v^2}}}{6}\)

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