If $n_1, n_2$ and $n_3$ are the fundamental frequencies of three segments into which a string is divided, then

Question

If $n_1, n_2$ and $n_3$ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency $n$ of the string is given by

Accepted Answer

1. **Identify the formula for fundamental frequency:** The fundamental frequency $n$ of a stretched string of length $l$, tension $T$, and linear mass density $\mu$ is given by $n = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$ .
2. **Relate length to frequency:** Assuming the tension $T$ and linear mass density $\mu$ remain constant for all segments of the string, the length $l$ is inversely proportional to the frequency $n$. Thus, $l = \frac{C}{n}$, where $C = \frac{1}{2}\sqrt{\frac{T}{\mu}}$ is a constant.
3. **Use the property of total length:** The total length of the string is equal to the sum of the lengths of the individual segments into which it is divided:
   $$l = l_1 + l_2 + l_3$$
4. **Substitute the length-frequency relationship:** Substituting $l = \frac{C}{n}$ into the length equation gives:
   $$\frac{C}{n} = \frac{C}{n_1} + \frac{C}{n_2} + \frac{C}{n_3}$$
   Dividing both sides by the constant $C$, we obtain the relationship between the fundamental frequencies:
   $$\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$$

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