When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of t

Question

When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $
u_1, 
u_2$ and $
u_3$ respectively. The original fundamental frequency ($
u$) of the string is:

Accepted Answer

1. **Identify the formula for fundamental frequency:** The fundamental frequency $
u$ of a stretched string of length $l$, tension $T$, and linear mass density $\mu$ is given by $
u = \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 $
u$. Thus, $l = \frac{C}{
u}$, 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 $l = \frac{C}{
u}$ into the length equation gives:
   $$\frac{C}{
u} = \frac{C}{
u_1} + \frac{C}{
u_2} + \frac{C}{
u_3}$$
   Dividing both sides by the constant $C$, we obtain the relationship between the fundamental frequencies:
   $$\frac{1}{
u} = \frac{1}{
u_1} + \frac{1}{
u_2} + \frac{1}{
u_3}$$

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