Longitudinal Waves on a Spring

Consider a horizontal coil spring with a horizontal pulse traveling along it. We will apply F = ma, and find the speed of the pulse. This is tricky. We are accustomed to force F applied to the end of the spring, and the spring constant k is the force/stretch distance (or compression distance). And the tension or compression is the same everywhere in the spring. The dynamic situation is different.

If you cut a spring in half, the spring constant will be double the original. (Think about it or experiment with it.) In general if the spring has length L and constant k, the spring constant associated with length D x is k' = kL/D x. This local value is what we need.

Let x represent the location of the “home” or equilibrium position of a particle of coil. Thus x varies from 0 to L.

Let s be the displacement of a particle of coil from its equilibrium position. In general, s is a function of x and t, and if the spring is in equilibrium, s = 0 everywhere. If F_o is the equilibrium tension and if s₁ is the displacement of at the beginning of interval D x and s₂ at the end of the interval, then the tension in that chunk of spring is F_o + k'(s₂ - s₁) = kL(s₂ - s₁)/D x . In the language of calculus, a small D is a d, and we say F = F_o + kLds/dx. When we recognize that s is a function of both x and t, and for F we need the change of s per change of only x, the d symbol changes to ¶ . We call ds/dx a derivative and ¶ s/¶ x a partial derivative.

Now let us do SF = ma. Let m = mass/length when the spring is in equilibrium. Consider a chunk D x. The net force is the difference between the two tensions, so F_o drops out and we have kL times the change of ¶ s/¶ x and the mass is m D x. We divide by D x and consider a very small
D x, denoted by ¶ x (we don't really need to know about taking the limit and all that). The change of
¶ s/¶ x divided by ¶ x is symbolized by ¶ ²s/¶ x². So we have
kL¶ ²s/¶ x² = m ¶ ²s/¶ t² (because acceleration is ¶ ²s/¶ t² ) and recalling m = m/L, and rearranging, we have

kL²/m¶ ²s/¶ x² = ¶ ²s/¶ t². (Call this the main differential equation.)

We have some unknown function s, a function of x and t. Whenever it happens that the acceleration is proportional to the change of slope per distance, as is the case here, we have a wave and the square of the velocity is the constant of proportionality. So

v = L(k/m)^1/2. If you don't buy the above, read below.

The plane wave s = Asin(Kx - w t) has ¶ ²s/¶ x² = -K² Asin(Kx - w t) and
¶ ²s/¶ t² = -w ²Asin(Kx-w t) (if you don't know this, take a calculus course to find out why) and v is w /K, because v = fl , f is w /2p , and l is K/2p . K is called the angular wave number, and it is not to be confused with the spring constant k. So combine these things with the main differential equation and show that
K²kL²/m = w ² and if you solve this for w /K, which is v, we find that
v = L(k/m)^1/2, as advertised.

Note that since v is proportional to the length, the time it takes a pulse to go the length of the spring, L/v, is independent of the length. Stretch it out farther and it goes faster in such a way as to make the time of travel the same.

Now oscillate back to the main junk on waves & stuff.

Other main pages: mechanics, fluid/heat/e&m, quantum, and alphabetized stuff in index.

Comments, questions: email fredrick.gram @tri-c.edu (remove space before @)