Q. 173.9( 17 Votes )

# The frequency of

Answer :

The frequency depends on the length L, mass per unit length m and tension F

f L^{a}F^{b}m^{c}

Let us assume a dimensionless constant k such that

f = k L^{a}F^{b}m^{c}

Dimensions of f = [T^{-1}]

Dimensions of L = [L]

Dimensions of F = [MLT^{-2}]

Dimensions of m = [ML^{-1}]

Dimensionally

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

= [M^{b+c} L^{a+b-c} T^{-2b}]

On comparing we can see that

b + c =0 …(i)

a + b – c =0 …(ii)

-2b = -1 …(iii)

On solving (i), (ii) and (iii)

a = -1, b = 1/2 and c = -1/2

So frequency can be expressed as

f = k L^{-1} F^{1/2} m^{-1/2}

f =

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