Problem 9 on Units & Dimensions

Description:

Problem

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

Solution

According to the problem,

f ∝ F^aL^bm^c

Introducing constant of proportionality,

f = K × F^aL^bm^c

The rules of dimensional analysis suggest that dimensions of both L.H.S. and R.H.S. must be equal to each other. Therefore,

Dimensions of L.H.S, (frequency = 1/Time Period),

1/[T]

Dimensions of R.H.S = ([M][L][T]^-2)^a [L]^b([M]/[L])^c

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

Equating dimensions on both sides,

[T]^-1 = [M]^a+c[L]^a+b[T]^-2a

The equations we obtain by comparing the exponents are,

a + c = 0

a + b - c = 0

-2a = -1

Results obtained are −

• a = 1/2

• b = -1

• c = -1/2

Hence, our final equation becomes,

f = K × √F × 1/L × 1/√m

Rearranging,

f = K/L√F/m