16.3 Wave speed on a stretched string  (Page 2/4)

The net force is on the small mass element can be written as

Using Newton’s second law, the net force is equal to the mass times the acceleration. The linear density of the string $\mu$ is the mass per length of the string, and the mass of the portion of the string is $\mu \text{Δ}x$ ,

Dividing by $F_T\text{Δ}x$ and taking the limit as $\text{Δ}x$ approaches zero,

Recall that the linear wave equation is

Therefore,

Solving for v , we see that the speed of the wave on a string depends on the tension and the linear density.

The wave speed of a guitar spring

Strategy

1. The speed of the wave can be found from the linear density and the tension $v=\sqrt{\frac{F_T}{\mu }}.$
2. From the equation $v=\sqrt{\frac{F_T}{\mu }},$ if the linear density is increased by a factor of almost 20, the tension would need to be increased by a factor of 20.
3. Knowing the velocity and the linear density, the velocity equation can be solved for the force of tension $F_T=\mu v^2.$

Solution

1. Use the velocity equation to find the speed:
   
   $v=\sqrt{\frac{F_T}{\mu }}=\sqrt{\frac{56.40\text{N}}{3.09\times {10}^{\mathrm{-4}}\text{kg/m}}}=427.23\text{m/s}.$
2. The tension would need to be increased by a factor of approximately 20. The tension would be slightly less than 1128 N.
3. Use the velocity equation to find the actual tension:
   
   $F_T=\mu v^2=5.78\times {10}^{\mathrm{-3}}\text{kg}\text{/}\text{m}{\left(427.23\text{m/s}\right)}^2=1055.00\text{N}.$
   
   This solution is within $7\text{\%}$ of the approximation.

Significance