Two identical piano wires have a fundamental frequency of 600Hzwhen kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of 6.0beats/s when both wires oscillate simultaneously?

We can find the frequency of the wire after tension is applied from the fundamental frequency and beat frequency using the formula for beat frequency. Then, using the formula for frequency, we can find the ratio of frequencies of the two wires. Then, rearranging it will give a fractional increase in the tension.

Formula:

The resonant frequency of the body in SHM,

The frequency difference or the beat frequency,

$\mathbf{∆}\mathit{f}\mathbf{=}{\mathit{f}}_{\mathbf{2}}\mathbf{-}{\mathit{f}}_{\mathbf{1}}$

…(2)

From equation (i), we can get that if we increase the string tension, then the frequency of the sound on the string will increase. Hence, the the change frequency can be given using equation (ii) as:

$f_2=f_1+∆f$

Substitute all the value in the above equation.

$\begin{array}{rcl}{f}_2& =& f_1+∆f\\ f_2& =& 600Hz+6Hz\\ & =& 606Hz\end{array}$

Now, using equation (i), we can write the old and new frequency as:

$f_1=\frac{1}{12}\sqrt{\frac{\tau }{\mu }}and{f}_2=\frac{1}{2L}\sqrt{\frac{\tau +∆\tau }{\mu }},$

So, the increase in the tension can be given as:

$\begin{array}{rcl}\frac{f_2}{f_1}& =& \sqrt{\frac{\tau +∆\tau }{\tau }}\\ & =& \sqrt{1+\left(\frac{∆\tau }{\tau }\right)}\\ \frac{∆\tau }{\tau }& =& {\left(\frac{f_2}{f_1}\right)}^2-1\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\frac{∆\tau }{\tau }& =& {\left(\frac{606Hz}{600Hz}\right)}^2-1\\ & =& 0.020\end{array}$

Hence, the fractional increase in the tension is, 0.020.