Physics - Sound Wave Question with Solution | TestHub

Let the frequency of the tuning fork be $f$.

The frequency of a stretched wire is directly proportional to the square root of its tension, $\nu \propto \sqrt{T}$.

Let ${\nu }_1$ and ${\nu }_2$ be the frequencies of the wire at tensions $T_1=225$ N and $T_2=256$ N respectively.

$\frac{{\nu }_1}{{\nu }_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{225}{256}}=\frac{15}{16}$

Since the number of beats is 6 in both cases and ${\nu }_2>{\nu }_1$, the frequency of the tuning fork must lie between ${\nu }_1$ and ${\nu }_2$.

Thus, ${\nu }_1=f-6$ and ${\nu }_2=f+6$.

Substituting these into the ratio:

$\frac{f-6}{f+6}=\frac{15}{16}$

$16(f-6)=15(f+6)$

$16f-96=15f+90$

$f=186$ Hz

Answer: 186 Hz