Physics - Wave on String Question with Solution | TestHub

Given ${\lambda }_1=50\text{ cm}=0.5\text{ m}$ and ${\lambda }_2=50.5\text{ cm}=0.505\text{ m}$.

The frequencies of the two waves are $f_1=\frac{v}{{\lambda }_1}$ and $f_2=\frac{v}{{\lambda }_2}$, where $v$ is the velocity of sound.

Given the difference in frequencies is $6\text{ Hz}$:

$f_1-f_2=6$

$\frac{v}{0.5}-\frac{v}{0.505}=6$

$v\left(\frac{1}{0.5}-\frac{1}{0.505}\right)=6$

$v\left(2-\frac{200}{101}\right)=6$

$v\left(\frac{202-200}{101}\right)=6$

$v\left(\frac{2}{101}\right)=6$

$v=\frac{6\times 101}{2}=303\text{ m/s}$

Answer: $303\text{ m/s}$