Transverse Wave Problem

1. Nov 20, 2005

Ginny Mac

Here is a transverse wave problem:

Consider transverse waves moving along a stretched spring fixed to an oscillator. Suppose the mass of the spring is 10 grams, its relaxed (not stretched) length is 1.00 meters, and its spring constant (if stretched) is 5.0 N/cm. The spring is tensioned via a pulley wheel and a hanging weight. The final length of the stretched spring is 1.50 meters (from the oscillator to the weight). How fast will transverse waves propogate along the stretched spring?

I am using v= square root of:(tension over mass density)
Initial mass density: 0.01 kg/1.00 m = 0.01 kg/m
Final mass density: 0.01 kg/1.50 m = 0.0067 kg/m
T = mass (weight) * gravity = x kg*9.8 m/s^2

So now to find T....
I have tried F = -kx, where mg= -kx, and T=-F. Therefore, T = (-500 N/m * 1.50 m) = 750 N, or kgm/s^2.
So... v = square root of: (750 N/0.0067 kg/m) = 334.6 m/s

I am just a little unsure. Could somebody please comment, whether right or wrong? If wrong, would you mind pointing me in the right direction? Thank you - I appreciate it very much.

2. Nov 21, 2005

mezarashi

The tension in the spring would be T = -kx, where k = 5N/cm and x = 50cm, as there is zero tension when the spring is at 100cm.

You should apply the solution to the differential equation, which has

$$\beta = \frac{\omega}{V}$$
$$V = \sqrt{\frac{T}{\rho}}$$

where V is the transverse wave velocity, T is the tension in the string, and rho is the linear mass density (mass/length). Using that formula, my answer differs.

The derivations here are quite lengthy, involving elastic theory + the wave equation, but I'll review if you'd like.

3. Nov 21, 2005

Ginny Mac

If you have a moment, would you mind reviewing this? I appreciate it. Here is my dilemma this semester : I am confused as to how I ended up in a Calc-based Physics class (eek!) but I am trudging along best I can! Thank you very much for your help -
Ginny

4. Nov 21, 2005

mezarashi

The first step is to derive the wave equation from the case of the classical string. In the case of transverse waves on the string, we see that the mass particles move up and down in the y direction, thus we will look at the forces involved in the y-direction.
For any small differential element $$\delta x$$ we can resolve the forces acting in the y-direction as the tension acting at both sides at different angles.

Force in y = T1 x sin a - T2 x sin b
But we also know that the horizontal tension T can be expressed both as T1cos a = T2cos b. Using this relationship we can simplify the equation to:

Force in y = T (tan a - tan b)

We also notice that tan a is the slope at any point x. tan b is the slope at any point x + a differential element dx, such that tan a = dy/dx at x and tan b = dy/dx at x+dx.

Applying Newton's law: F = ma or $$F = m\frac{d^2 y}{dt^2}$$
$$F = \rho \delta x\frac{d^2 y}{dt^2} = T (dy/dx_1 + dy/dx_2)$$
where rho represents the linear mass density. As the differential element approaches zero, our equation becomes.

$$\frac{\rho}{T} \frac{d^2 y}{dt^2} = \frac{d^2 y}{dx^2}$$
which is our wave equation (D'Alambert's wave equation in 1-dimension) if we substitute $$c^2 = \frac{T}{\rho}$$.

Now you can use theories from your study of differential equations to solve, so you have the solution to be in the possible form of:

$$y = A exp(j(\omega t - \beta x)) + B exp(j(\omega t + \beta x))$$
Properties of the wave equation tells us that $$\beta = \frac{\omega}{v}$$ where $$v = c = \sqrt{\frac{T}{\rho}}$$

I personally believe that the understanding of the solution to the wave problem and the meaning of the terms in the final wave equation solution is the most important part of all. I remember such classes being tough too >.< Good luck!

Last edited: Nov 21, 2005