Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $\nu_1, \nu_2$ and $\nu_3$ respectively. The original fundamental frequency ( $\nu$ ) of the string is:

$\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}$

$\nu=\nu_1+\nu_2+\nu_3$

$\frac{1}{\nu}=\frac{1}{\nu_1}+\frac{1}{\nu_2}+\frac{1}{\nu_3}$

$\frac{1}{\sqrt{\nu}}=\frac{1}{\sqrt{\nu_1}}+\frac{1}{\sqrt{\nu_2}}+\frac{1}{\sqrt{\nu_3}}$

Step-by-Step Solution

Identify the formula for fundamental frequency: The fundamental frequency $\nu$ of a stretched string of length $l$ , tension $T$ , and linear mass density $\mu$ is given by $\nu = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$ .
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 $\nu$ . Thus, $l = \frac{C}{\nu}$ , where $C = \frac{1}{2}\sqrt{\frac{T}{\mu}}$ is a constant.
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: $l = l_1 + l_2 + l_3$
Substitute the length-frequency relationship: Substituting $l = \frac{C}{\nu}$ into the length equation gives: $\frac{C}{\nu} = \frac{C}{\nu_1} + \frac{C}{\nu_2} + \frac{C}{\nu_3}$ Dividing both sides by the constant $C$ , we obtain the relationship between the fundamental frequencies: $\frac{1}{\nu} = \frac{1}{\nu_1} + \frac{1}{\nu_2} + \frac{1}{\nu_3}$

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