Two tuning forks A and B produce 10 beats per second when sounded together. On slightly loading fork A with a little wax, it was observed that 15 beats are heard per second. If the frequency of fork B is 480 Hz. what is the frequency of fork A (in Hz) before it was loaded?

Let v₁ be the frequency of A and v₂ that of fork B.
There are two possibilities (a) v₁ > v₂ and (b) v₁ < v₂.

$\text{ (a) }{\mathrm{v}}_1>{\mathrm{v}}_2:{\mathrm{v}}_2=480\mathrm{Hz},{\mathrm{v}}_{\mathrm{b}}=10\mathrm{Hz}$

Now, ${\mathrm{v}}_{\mathrm{b}}={\mathrm{v}}_1-{\mathrm{v}}_2$

$\therefore {\mathrm{v}}_1={\mathrm{v}}_2+{\mathrm{v}}_{\mathrm{b}}=480+10=490\mathrm{Hz}$

On loading fork A its frequency decreases. In other words, v₁ decreases.

$\therefore {\mathrm{v}}_1-{\mathrm{v}}_2={\mathrm{v}}_{\mathrm{b}}$ should also decrease. But v_b increase to 15 beats/s. Hence v₁ is not greater than v₂.

$\text{ (b) }{\mathrm{v}}_1<{\mathrm{v}}_2:\text{ In this case }{\mathrm{v}}_2-{\mathrm{v}}_1={\mathrm{v}}_{\mathrm{b}}\text{. }$

${\mathrm{v}}_1={\mathrm{v}}_2-{\mathrm{v}}_{\mathrm{b}}=480-10=470 \mathrm{Hz}$