Answer

Verified

54k+ views

**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.

Recently Updated Pages

Why do we see inverted images in a spoon class 10 physics JEE_Main

A farsighted man who has lost his spectacles reads class 10 physics JEE_Main

State whether true or false The outermost layer of class 10 physics JEE_Main

Which is not the correct advantage of parallel combination class 10 physics JEE_Main

State two factors upon which the heat absorbed by a class 10 physics JEE_Main

What will be the halflife of a first order reaction class 12 chemistry JEE_Main

Other Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Which of the following distance time graph is representing class 11 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

Choose the correct statements A A dimensionally correct class 11 physics JEE_Main

The figure below shows regular hexagons with charges class 12 physics JEE_Main