$A$ tuning fork of known frequency $256\,Hz$ produces $5$ beats per second with the vibrating string of a guitar. The beat frequency decreases to $2$ beats per second when the tension in the guitar string is slightly increased. The frequency of the guitar string before increasing the tension was ..... $Hz$.

$A$ set of $28$ tuning forks is arranged in an increasing order of frequencies. Each fork produces '$x$' beats per second with the preceding fork and the last fork is an octave of the first. If the frequency of the $12^{\text{th}}$ fork is $152 \text{ Hz}$,the value of '$x$' (number of beats per second) is:

Two identical straight wires are stretched so as to produce $6$ beats per second when vibrating simultaneously. On changing the tension in one of them,the beat frequency remains unchanged. Denoting by ${T_1}$ and ${T_2}$ the higher and the lower initial tensions in the strings,respectively,then it could be said that while making the above change in tension:

The maximum number of beats per second that a human ear can distinguish is

$A$ string under a tension of $129.6 \ N$ produces $10 \ beats/s$ when it is vibrated along with a tuning fork. When the tension in the string is increased to $160 \ N$,it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork in $Hz$.

Two vibrating tuning forks produce progressive waves given by $Y_1 = 4\sin(500\pi t)$ and $Y_2 = 2\sin(506\pi t)$. The number of beats produced per minute is: