Compression Pulse in a Massive Spring
Solution
Treat the massive spring as a continuous elastic medium.
For the complete spring,
\[ k=\frac{F}{\Delta l}. \]Consider a small element of the spring having natural length \(dx\). Since the spring constant is inversely proportional to its length,
\[ k_{dx}=k\frac{l}{dx}. \]Let \(u(x,t)\) be the longitudinal displacement of a point of the spring. The extension of the small element \(dx\) is approximately
\[ d(\Delta l)=\frac{\partial u}{\partial x}\,dx. \]Therefore, the force transmitted through the element is
\[ F=k_{dx}\,d(\Delta l). \] \[ F= \frac{kl}{dx} \left( \frac{\partial u}{\partial x}dx \right) = kl\frac{\partial u}{\partial x}. \]Hence, for an element \(dx\), the net force is
\[ dF= kl\frac{\partial^2u}{\partial x^2}dx. \]Since the spring is uniform, its linear mass density is
\[ \lambda=\frac{m}{l}. \]Therefore, the mass of the element \(dx\) is
\[ dm=\frac{m}{l}dx. \]Applying Newton’s second law,
\[ \frac{m}{l}dx \frac{\partial^2u}{\partial t^2} = kl\,dx \frac{\partial^2u}{\partial x^2}. \]Cancelling \(dx\),
\[ \frac{\partial^2u}{\partial t^2} = \frac{kl^2}{m} \frac{\partial^2u}{\partial x^2}. \]Comparing this with the standard one-dimensional wave equation,
\[ \frac{\partial^2u}{\partial t^2} = v^2 \frac{\partial^2u}{\partial x^2}, \]we obtain the wave speed
\[ \boxed{ v=l\sqrt{\frac{k}{m}} }. \]The pulse has to travel a distance \(l\). Therefore,
\[ t=\frac{l}{v}. \] \[ t= \frac{l}{ l\sqrt{k/m} }. \]Interesting result: the time taken by the pulse is independent of the length of the spring, provided \(m\) and \(k\) refer to the mass and force constant of the complete spring.