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The frequency of vibration of a string depends on the length L between the nodes, the tension F in the string and its mass per unit length m. Guess the expression for its frequency from dimensional analysis.

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#### Solution

Frequency, f \[\propto\] L^{a}F^{b}m^{c}

f = kL^{a}F^{b}m^{c} ...(1)

Dimension of [f] = [T^{−1}]

Dimension of the right side components:

[L] = [L]

[F] = [MLT^{−2}]

[m] = [ML^{−1}]

Writing equation (1) in dimensional form, we get:

[T^{−1}] = [L]^{a} [MLT^{−2}]^{b} [ML^{−1}]^{c}

[M^{0}L^{0}T^{−1}] = [M^{b + c }L^{a + b − c} T^{−2b}]

Equating the dimensions of both sides, we get:

b + c = 0 ....(i)

a + b −c = 0 ....(ii)

−2b = −1 ....(iii)

Solving equations (i), (ii) and (iii), we get:

∴ Frequency, f = kL

^{−1}F

^{1}

^{/}

^{2}m

^{−1}

^{/2}

\[= \frac{k}{L} F^{1/2} m^\left( - \frac{1}{2} \right) = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}\]

\[f = \frac{k}{L} \sqrt{\left( \frac{F}{m} \right)}\]

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