Wave speed of a stretched string

[Aaron7]

Homework Statement

A rubber string when unstretched has length L₀ and mass per unit length μ₀. It is clamped by its ends and stretched by ΔL. The tension is T=κΔL / L₀.

Show that the wave speed on the rubber string when stretched by ΔL is

1/L₀ √( (κ/μ₀) ΔL(L₀+ΔL) )

Homework Equations

c = √(T/μ)
δ²y/δx² = 1/c² δ²y/δt²

The Attempt at a Solution

Using the formula c = √(T/μ) and putting in T I get:

c = √( κΔL/L₀μ₀ )

but I am not sure how to arrive at the answer or whether this is correct?

Many thanks.

 

[anathema]

Your attempt at solving the problem is on the right track. Here is a step-by-step solution to help you arrive at the correct answer:

1. Start by writing down the general formula for wave speed: c = √(T/μ). This equation relates the wave speed (c) to the tension (T) and the mass per unit length (μ) of the string.

2. Next, we need to substitute the given values into this equation. We know that the tension (T) is equal to κΔL/L0, so we can replace T in the formula with this expression. This gives us: c = √( (κΔL/L0)/μ).

3. We also know that the mass per unit length (μ) of the string is μ0, so we can substitute this value into the equation. This gives us: c = √( (κΔL/L0)/μ0).

4. Now, we can simplify the expression in the square root by multiplying the top and bottom by L0. This gives us: c = √( (κΔL)/ (L0μ0)).

5. We can further simplify this expression by noticing that the numerator (κΔL) is equal to κΔL(L0+ΔL) - L0κΔL. This means that we can rewrite the expression as: c = √( (κΔL(L0+ΔL) - L0κΔL)/ (L0μ0)).

6. Now, we can factor out κΔL from the numerator, which gives us: c = √( κΔL (L0+ΔL - L0)/ (L0μ0)).

7. Simplifying the expression in the parentheses, we get: c = √( (κΔLΔL)/ (L0μ0)).

8. Finally, we can factor out ΔL from the square root, which gives us the final expression: c = √( (κ/μ0) ΔL(L0+ΔL) )/L0.

This matches the given answer: 1/L0 √( (κ/μ0) ΔL(L0+ΔL) ).

I hope this helps clarify the steps needed to arrive at the correct answer. Keep up