Two tuning forks have frequencies $380 \ Hz$ and $384 \ Hz$ respectively. When they are sounded together,they produce $4$ beats per second. After hearing the maximum sound,how long will it take to hear the minimum sound in seconds?

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

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 $\rightarrow T_1 > T_2$)

When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously,$4$ beats per second are heard. Now,some tape is attached to the prong of fork $2$. When the tuning forks are sounded again,$6$ beats per second are heard. If the frequency of fork $1$ is $200 \, Hz$,then what was the original frequency of fork $2$ (in $, Hz$)?

If two progressive sound waves represented by $y_1 = 3 \sin(250 \pi t)$ and $y_2 = 2 \sin(260 \pi t)$ (where displacement is in metre and time is in second) superimpose,then the time interval between two successive maximum intensities is (in $s$)

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: