Two identical piano wires have a fundamental frequency of $600 \text{ Hz}$ when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of $6$ beats per second when both wires vibrate simultaneously?

Two tuning forks $A$ and $B$ are sounded together and produce $6$ beats per second. The frequency of tuning fork $B$ is $384 \,Hz$. When tuning fork $A$ is loaded with wax,the number of beats per second becomes $4$. What is the frequency of tuning fork $A$ in $Hz$?

Two identical straight wires are stretched so as to produce $6$ beats per second when vibrating simultaneously with tensions $T_1$ and $T_2$ respectively. On changing the tension slightly in one of them,the beat frequency remains unchanged. This will happen when (Given $\rightarrow T_1 > T_2$)

Two waves of lengths $50 \; cm$ and $51 \; cm$ produce $12$ beats per second. The velocity of sound is .... $m/s$.

$A$ tuning fork of frequency $400 \ Hz$ is in unison with a sonometer wire. The number of beats heard per second,when the tension in the wire is increased by $1 \%$,will be

The lengths of the two pipes open at both ends are $L$ and $(L + L_1)$. If they are sounded together, the beat frequency will be ($v$ = velocity of sound in air)