Answer
$7.06Hz$
Work Step by Step
We know that
$f_n=\frac{n}{2L}\sqrt{\frac{F}{\mu}}$
We are given that tension increases by $2.25\%$
$\implies f_n^{\prime}=\frac{n}{2L}\sqrt{\frac{1.0225F}{\mu}}=\sqrt{1.0225}f_n$
Now we can find the beat frequency as
$f_{beat}=f_n^{\prime}-f_n=(\sqrt{1.0225-1})(631)=7.06Hz$
