$A$ person standing at a distance of $6\, m$ from a source of sound receives sound waves in two ways: one directly from the source and the other after reflection from a rigid boundary,as shown in the figure. The maximum wavelength for which the person will receive maximum sound intensity is .... $m$.

Three waves of equal frequency having amplitudes $10 \,\mu m, 4 \,\mu m$ and $7 \,\mu m$ arrive at a given point with successive phase difference of $\frac{\pi}{2}$. The amplitude of the resulting wave in $\mu m$ is given by

Two sources of sound $A$ and $B$ produce waves of $350 Hz$,and they vibrate in the same phase. $A$ particle $P$ is vibrating under the influence of these two waves. If the amplitudes at point $P$ produced by the two waves are $0.3 mm$ and $0.4 mm$,what will be the resultant amplitude of the point $P$ when $AP - BP = 25 cm$ and the velocity of sound is $350 m/s$?

Two sound waves each of wavelength $\lambda$ and having the same amplitude $A$ from two sources $S_1$ and $S_2$ interfere at a point $P$. If the path difference $S_2P - S_1P = \lambda/3$,then the amplitude of the resultant wave at point $P$ will be $[\cos(120^{\circ}) = -0.5]$.

Two sinusoidal waves given below are superposed:
$y_1 = A \sin \left(kx - \omega t + \frac{\pi}{6}\right), \quad y_2 = A \sin \left(kx - \omega t - \frac{\pi}{6}\right)$
The equation of the resultant wave is:

Two interfering waves have the same wavelength,frequency,and amplitude. They are traveling in the same direction but are $90^o$ out of phase. Compared to the individual waves,the resultant wave will have the same.