Looking For A Real Spring

This problem departs from the oscillating spring in that the material of the spring has mass. Instead of propelling a mass at the end of the spring through a large motion, the driving force from the compressed/extended spring is used to accelerate and decelerate mass that is—on average—closer to the fixed end, and thus, requires less energy. Therefore, we should expect the massive spring to vibrate with a frequency faster than the massless spring with a massive weight at the end.

It would be cumbersome to approach this problem using Newton's second law as we need to consider the distribution of mass along the length of the spring—not just the bob at the end like usual. Therefore, we take the Lagrangian approach which makes it easier to deal with distributed quantities like the many particles that make up the spring.

If the particles of the spring vibrated with a uniform velocity, then we could simply write the kinetic energy of the spring as $\frac12 mv^2$ . However, when the spring vibrates, the particles near the fixed end oscillate with a very small amplitude while the particles near the free end have a bigger amplitude. In particular, if the free end vibrates with amplitude $A$ , a particle halfway along the spring will have an amplitude, and therefore a speed half as large. Thus, if the end vibrates with speed $\dot{s}$ , a particle a distance $l$ along the spring will have the speed $u(l)=\frac{l}{L}\dot{s}$ , and kinetic energy $\delta T = \frac12 u(l)^2 \sigma dl$ .

As the mass of the spring is evenly distributed, we can write a mass density $\sigma = M/L$ so that the mass of a small segment of the spring of length $dl$ is given by $\sigma dl$ .

To get the kinetic energy $T$ of the entire spring, we can simply add up the kinetic energy of the small mass elements, $T = \int \delta T = \int \frac12 u(l)^2 \sigma dl$ . We find

$\begin{aligned} T &= \frac{\sigma}{2} \int\limits_0^L u(l)^2 dl \\ &= \frac{\sigma}{2L^2}\dot{s}^2\int\limits_0^L l^2 dl \\ &= \frac{M\dot{s}^2}{6} \end{aligned}$

The potential energy of the system is given by the Hooke potential $U = \frac12 ks^2$ so that the Lagrangian is

$\begin{aligned} \mathcal{L} &= T - U \\ &= \frac{M\dot{s}^2}{6} - \frac12 ks^2 \end{aligned}$

Taking the derivatives—according to the Euler-Lagrange equations—we find

$\begin{aligned} \frac{d}{dt}\frac{d\mathcal{L}}{d\dot{s}} &= \frac{d \mathcal{L}}{ds} \\ -\frac{M}{3} \ddot{s} &= ks \\ \ddot{s} &= -\overbrace{\frac{3k}{M}}^{\omega^2} s \end{aligned}$

Which is the expression for a harmonic oscillator with frequency $\omega = \sqrt{\frac{3k}{M}}$ .

First write the energy of the spring during motion i.e. after the spring is released.

The spring will possess elastic potential energy and kinetic energy.

We know that when one end of a spring with mass is fixed its kinetic energy= $\frac{1}{6}$ M $v^{2}$

Its elastic potential energy = $\frac{1}{2}$ k $x^{2}$

Total Energy ,

$\frac{1}{6}$ M $v^{2}$ + $\frac{1}{2}$ k $x^{2}$ = E

On differentiating w.r.t time on both sides,

$kxv$ + $\frac{Mva}{3}$ = $0$

$a$ = - $\frac{3kx}{M}$

Comparing to the equation, $a$ = - $ω^{2}$ $x$

$ω$ = $\sqrt{\frac{3k}{M}}$

Substituting the given values,

$ω$ = $\sqrt{3}$

= $1.732$