Fundamental frequency if string halved and tension * 4

[moenste]

Homework Statement

The fundamental frequency of vibration of a particular string is f. What would the fundamental frequency be if the length of the string were to be halved and the tension in it were to be increased by a factor of 4?

Answer: 4 f

2. The attempt at a solution
We have f = f₁, 0.5 L and 4 T.

Substitute in: f_n = (n / 2 L) * (√ T / μ)
f₁ = (1 / 2 * 0.5 L) * (√ 4 T / μ)
f₁ = (2 √ T / μ) / L
f₁ = (2 v) / L
f₁ L = 2 v
In general: f₁ = v / (2 L) so
v / 2 = 2 v
v = 4 v
f₁ = v / (2 L) so
v = 2 f₁ L so
2 f₁ L = 8 f₁ L
f₁ = 4 f₁
Because f = f₁ then
f = 4 f

But I am not sure whether this logic is right or not. Any help please?

 

[gTurner]

Yes, that is correct. One way to think about this without all the math is, if you cut the length of the rope in half, but the velocity (due to the tension and mass/unit length) stays constant, then the frequency should double since freq*wavelength = velocity. Then, if you quadruple the tension, because the velocity is dependent on the (square) root of the tension, this causes the velocity to increase by a factor of 2, and thus causing the frequency to double since the wavelength is fixed. Therefore, you've essentially doubled the frequency twice; once by cutting the string in half, and then again by quadrupling the tension. Hope that helps.