Question

When two identical wires on a sonometer are kept under same tension, their fundamental frequency is $500Hz$. In order to produce five beats per second, the percentage change in the tension of one of the wires will be:

• A. $2\text{\%}$
• B. $4\text{\%}$
• C. $6\text{\%}$
• D. $8\text{\%}$

Answer 1

The correct option is A $2\text{\%}$

$\nu =\frac{1}{2l}\sqrt{\frac{T}{m}}$ where $\nu$ is frequency, $l$ is the length of the string, $T$ is the tension on the string and m is mass per unit length.

Let's assume, for tension $T_1$ frequency is ${\nu }_1$ and for tension

$T_2$ frequency is ${\nu }_2$.

$\frac{{\nu }_2}{{\nu }_1}=\sqrt{\frac{T_2}{T_1}}$

$1-\frac{{\nu }_2^2}{{\nu }_1^2}=1-\frac{T_2}{T_1}$

Now use $|{\nu }_2-{\nu }_1|=\text{number of beats per second}$

$1-\frac{(500-5{)}^2}{{500}^2}=\frac{T_1-T_2}{T_1}$

$1-(1-\frac{1}{100}{)}^2=\frac{T_1-T_2}{T_1}$

$1-(1-\frac{2}{100})=\frac{T_1-T_2}{T_1}$

$\frac{2}{100}\times 100=\frac{T_1-T_2}{T_1}\times 100\%$

Percentage change in tension = $2$.