Chapter 11, Problem 58P

To determine

To Find: The percentage change in the tension in the string.

Answer to Problem 58P

The percentage change in the tension in the string is decrease up to $4.8\text{ \%}$ to correct the frequency.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Measured frequency }{f}_1=205\text{ Hz}\\ \text{Original frequency }{f}_2=200\text{ Hz}\end{array}$

Formula used:

Speed of the wave on string is calculated by following formula,

  $v=\sqrt{\frac{T}{\left(\frac{m}{L}\right)}}$

Here,

  $T$ is string tension,

  $m$ is string mass,

  $L$ is string length.

The fundamental frequency of string with both end fixed is calculated by following formula,

  $f_n=\frac{n\times v}{2\times L}$

Here $f_n$ is the frequency of $n-1$ overtone,

  $v$ is speed of wave,

  $l$ is the length of vibrating string.

  $n$ = no. of fixed end - $1$ .

Calculation:

Ratio of original frequency to measured frequency is required to find the percentage change in tension. It can be described as follows,

  $\begin{array}{l}\frac{f_2}{f_1}=\sqrt{\frac{T_2}{T_1}}\\ T_2=T_1\times {\left(\frac{f_2}{f_1}\right)}^2\\ T_2=T_1\times {\left(\frac{200}{205}\right)}^2\\ T_2=T_1\times 0.952\\ \text{So, }{T}_2=95.2\%\text{ of }{T}_1\end{array}$

So, the tension in the string should be decreased by $100\%-95.2\%=4.8\%$ .

Conclusion:

Thus, the percentage change in the tension in the string is decrease up to $4.8\%$ to correct the frequency.