First slide

Dimensions of physical quantities

Question

The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string and the mass per unit length $μ$ of string. Derive the formula for frequency.

Moderate

A

$f = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$
B

$f = \frac{1}{2 l} \sqrt{\frac{L}{μ}}$
C

$f = \frac{1}{2 l} \sqrt{\frac{M}{μ}}$
D

$f = \frac{1}{3 l} \sqrt{\frac{L}{μ}}$

Solution

Suppose, that the frequency f depends on the tension raised to the power a, length raised to power b and mass per unit length raised to power c. Then,
$f \propto F^{a} l^{b} μ^{c} o r, f = k F^{a} l^{b} μ^{c}$
Here, k is a dimensionless constant. Thus,
$[f] = [F]^{a} [η]^{b} [μ]^{c} o r, [M^{0} L^{0} T^{- 1}] = {[{MLT}^{- 2}]}^{a} [L]^{b} {[{ML}^{- 1}]}^{c} o r, [M^{0} L^{0} T^{- 1}] = [M^{a + c} L^{a + b - c} T^{- 2 a}]$
For dimensional balance, the dimensions on both sides should be same.
Thus, $a + c = 0 \dots (ii)$
$a + b - c = 0 . . . (i i i) a n d - 2 a = - 1 \dots (iv)$
Solving these three equations, we get
$a = \frac{1}{2}, c = - \frac{1}{2} and b = - 1$
Substituting these values in Eq. (i), we get
$f = k (F)^{1 / 2} (l)^{- 1} (μ)^{- 1 / 2} o r, f = \frac{k}{l} \sqrt{\frac{F}{μ}}$
Experimentally, the value of k is found to be $\frac{1}{2}$
Hence,
$f = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$

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