# Question #85ba3

Aug 31, 2016

The fundamental frequency of a streched string is given by the follwing formula.

$n = \frac{1}{2 l} \sqrt{\frac{T}{m}}$

Where

$l \to \text{Length of the string}$

$T \to \text{Tension of the string}$

$m \to \text{Mass per unit length of the string}$

So T and m remaining constant

$n \propto \frac{1}{l}$

let the frequency of the string is ${n}_{1}$ when its length is 20cm and its frequency is ${n}_{2}$ when its length is 21cm.
Si by the law of length

${n}_{1} / {n}_{2} = \frac{21}{20}$

$L e t \text{ } {n}_{1} = 21 x H z \mathmr{and} {n}_{2} = 20 x H z$

Let the frequency of tuning fork be ${n}_{f}$ It forms 10 beats with both length of the sonometer string.

So ${n}_{1} > {n}_{f} > {n}_{2}$

So
${n}_{1} - {n}_{f} = 10 \implies 21 x - {n}_{f} = 10. \ldots . . \left(1\right)$

Again

${n}_{f} - {n}_{1} = 10 \implies {n}_{f} - 20 x = 10. \ldots . . \left(2\right)$

Adding (1) and (2) we get
$x = 20$

Inserting the value of x in (2)

${n}_{f} - 20 \cdot 20 = 10$

${n}_{f} = 410 H z$

$\text{Hence frequency of tuning fork} = 410 H z$