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**Hint:**Assume a proportional relation of \[f\] with \[F\], \[l\] and \[\mu \]. Then use the dimensional analysis to derive the formula.

**Complete step-by-step solution**

Let the frequency \[f\] be proportional to the tension \[F\] raised to the power \[x\], length \[l\] raised to the power \[y\] and mass per unit length \[\mu \] raised to the power \[z\], i.e.

\[f \propto {F^x}{l^y}{\mu ^z}\]

Removing the proportionality sign with the constant \[c\], we get

\[f = c({F^x}{l^y}{\mu ^z})\] …...(1)

For the above equation to be correct, the dimensions of the quantity in the LHS should be equal to the dimensions of the quantities in the RHS.

Replacing the quantities of the above equation with their dimensions, we get

\[\left[ f \right] = \left[ {{M^0}{L^0}{T^{ - 1}}} \right]\], \[\left[ F \right] = \left[ {{M^1}{L^1}{T^{ - 2}}} \right]\], \[\left[ l \right] = \left[ {{M^0}{L^1}{T^0}} \right]\]and \[\left[ \mu \right] = \left[ {{M^1}{L^{ - 1}}{T^0}} \right]\]

\[\because c\] is a constant, so it has no dimensions.

Substituting these in (1) we get

\[\left[ {{M^0}{L^0}{T^{ - 1}}} \right] = {\left[ {{M^1}{L^1}{T^{ - 2}}} \right]^x}{\left[ {{M^0}{L^1}{T^0}} \right]^y}{\left[ {{M^1}{L^{ - 1}}{T^0}} \right]^z}\]

\[\left[ {{M^0}{L^0}{T^{ - 1}}} \right] = \left[ {{M^x}{L^x}{T^{ - 2x}}} \right]\left[ {{M^0}{L^y}{T^0}} \right]\left[ {{M^z}{L^{ - z}}{T^0}} \right]\]

On simplifying, we get

\[\left[ {{M^0}{L^0}{T^{ - 1}}} \right] = \left[ {{M^{x + z}}{L^{x + y - z}}{T^{ - 2x}}} \right]\]

Comparing the exponents of similar dimensions, we get

\[x + z = 0\] ………..(2)

\[x + y - z = 0\] ………...(3)

And

\[ - 2x = - 1\] …………..(4)

From (4), we get \[x = \dfrac{1}{2}\]

Putting this in (2)

\[\dfrac{1}{2} + z = 0\]

\[z = - \dfrac{1}{2}\]

Putting the values of \[x,z\] in (3)

\[\dfrac{1}{2} + y - \left( { - \dfrac{1}{2}} \right) = 0\]

\[y + 1 = 0\]

Finally, \[y = - 1\]

\[\therefore x = \dfrac{1}{2}, y = - 1,z = - \dfrac{1}{2}\]

Putting these values in (1)

\[f = c({F^{\dfrac{1}{2}}}{l^{ - 1}}{\mu ^{ - \dfrac{1}{2}}})\]

Or, \[f = \dfrac{c}{l}\sqrt {\dfrac{F}{\mu }} \]

Hence, the formula for the frequency is

\[f = \dfrac{c}{l}\sqrt {\dfrac{F}{\mu }} \], where \[c\] is a constant.

Additional Information: The value of the c can be found experimentally. By experiment, it is found that \[c = \dfrac{1}{2}\]. Putting this value in the expression of frequency derived above, the final formula for frequency becomes:

\[f = \dfrac{1}{{2l}}\sqrt {\dfrac{F}{\mu }} \]

The formula derived above is used in finding the set of frequencies, called the normal modes of oscillation. The formula derived above is used to find the effect of increasing or decreasing the tension of a musical instrument on the frequency.

**Note:**While deriving a formula using dimensional analysis, be careful while writing the dimensions of each quantity. We can use any physical formula of each quantity to find its dimensions. Always prefer to use the formula which relates the quantity with more fundamental quantities.

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