Two tuning forks $A$ and $B$ produce notes of frequencies $258 \,Hz$ and $262 \,Hz$. An unknown note sounded with $A$ produces certain beats. When the same note is sounded with $B$, the beat frequency gets doubled. The unknown frequency is (in $\,Hz$)

The wavelengths of two notes in air are $\frac{36}{195} \,m$ and $\frac{36}{193} \,m$. Each note produces $10$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is

$A$ tuning fork $A$ produces $5$ beats per second with a tuning fork of frequency $480 \ Hz$. When a little wax is stuck to a prong of fork $A$,the number of beats heard per second becomes $2$. What is the frequency of tuning fork $A$ before the wax is stuck to it (in $Hz$)?

The ratio of intensities between two coherent sound sources is $4:1$. The difference of loudness in $dB$ between maximum and minimum intensities when they interfere in space is:

$A$ tuning fork of unknown frequency produces $3 \, \text{beats/sec}$ with a standard tuning fork of frequency $384 \, \text{Hz}$. The beat frequency decreases when a small piece of wax is placed on the prong of the unknown tuning fork. The frequency of the unknown tuning fork is .... $\text{Hz}$

$A$ tuning fork gives $3$ beats with $50 \ cm$ length of sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)