Two tuning forks $F_1$ and $F_2$ produce $6$ beats per second when sounded together. The frequency of $F_1$ is $256 \ Hz$. When wax is applied to $F_2$,the number of beats per second remains $6$. What is the frequency of $F_2$ in $Hz$?

Two strings $A$ and $B$ produce beats of frequency $\Delta f_1 > 0$. The tension in string $A$ is slightly increased and the beat frequency is found to be $\Delta f_2 > 0$. If the original frequency of $A$ is $f_0$ and $\Delta f_2 < \Delta f_1$,then the frequency of $B$ is

$A$ tuning fork of frequency $480 \ Hz$ produces $10$ beats per second when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before?

$16$ tuning forks are arranged in the order of increasing frequencies. Any two successive forks give $8$ beats per second when sounded together. If the frequency of the last fork is twice the first,then the frequency of the first fork is:

The air columns in two tubes closed at one end vibrating in their fundamental modes produce $2$ beats per second. The number of beats produced per second when the same tubes are vibrated in their fundamental mode with their both ends open are

$A$ tuning fork of frequency $392 \, Hz$ resonates with $50 \, cm$ length of a string under tension $T$. If the length of the string is decreased by $2 \%$,keeping the tension constant,the number of beats heard when the string and the tuning fork are made to vibrate simultaneously is