$A$ sound wave of wavelength $32 \ cm$ enters the tube at $S$ as shown in the figure. Then the smallest radius $r$ so that a minimum of sound is heard at detector $D$ is ... $cm$.

There is a destructive interference between two waves of wavelength $\lambda$ coming from two different paths at a point. To get maximum sound or constructive interference at that point,the path of one wave is to be increased by

Two sound waves having intensities $I$ and $4I$ interfere to produce an interference pattern. The phase difference between the waves is $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

Two simple harmonic waves having equal amplitudes of $8\,cm$ and equal frequency of $10\,Hz$ are moving along the same direction. The resultant amplitude is also $8\,cm$. The phase difference between the individual waves is $..................$ degree.

If two waves represented by $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \pi / 3)$ interfere at a point,find the amplitude of the resultant wave.

Two waves of amplitudes $A_1$ and $A_2$ respectively are superimposed. The ratio between the maximum and minimum intensities of the resultant waves is $9 : 4$. The value of $A_2 / A_1$ is [Assume $A_1 > A_2$]