$56$ tuning forks are arranged such that each fork produces $4$ beats per second with its previous one. If the frequency of the last fork is twice that of the first, the frequency of the $19^{\text{th}}$ fork is . . . . . . (in $\text{Hz}$)

The displacement at a point due to two waves are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$. The result due to their superposition will be

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384 \,Hz$. When one of the prongs of the fork $A$ is filed and sounded with $B$, the beat frequency increases. Then the frequency of the fork $A$ is: (in $\,Hz$)

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?

In case of beats,produced by the superposition of two waves of equal amplitudes,if maximum intensity is $x$ times the intensity of a superposing wave,then $x =$ ......

Two sound waves of wavelengths $\lambda_1$ and $\lambda_2$ $(\lambda_2 > \lambda_1)$ produce $n$ beats per second. The speed of sound is: