Mass per Unit Length of Violin Strings

[e(ho0n3]

[SOLVED] Mass per Unit Length of Violin Strings

Homework Statement

Each string on a violin is tuned to a frequency 1.5 times that of its neighbor. If all the strings are to be placed under the same tension, what must be the mass per unit length of each string relative to that of the lowest string?

Homework Equations

$$f = v/(2L)$$
$$v = \sqrt{T / \mu }$$

The Attempt at a Solution

Suppose the lowest string is tuned at the fundamental frequency f₁ = v₁/(2L), where v₁ is the velocity of the standing wave on the lowest string and L being the length of the string. For n > 1, f_n = 1.5^{n - 1}f₁ = v_n/(2L). So,

$$1.5^{n - 1}v_1 = v_n$$

Now, since $v_i = \sqrt{T/\mu_i}$, then

$$\mu_n = \frac{\mu_1}{1.5^{2(n-1)}}$$

Is that right?

 

[e(ho0n3]

I have checked the answer with the book. It coincides. Thanks anyways.

 

[m2003]

Yes, that is correct. The mass per unit length of each string relative to the lowest string will decrease as the frequency increases, following the relationship \mu_n = \frac{\mu_1}{1.5^{2(n-1)}}. This means that the higher strings will have a lower mass per unit length compared to the lowest string, in order to achieve the same tension and tuning. This is due to the fact that as the frequency increases, the wavelength decreases, requiring a lower mass per unit length to maintain the same tension and frequency. This relationship is important in understanding the physics behind the tuning and construction of violin strings.