Two coherent sources of sound,$S_{1}$ and $S_{2}$,produce sound waves of the same wavelength,$\lambda = 1\, m$,in phase. $S_{1}$ and $S_{2}$ are placed $1.5\, m$ apart (see figure). $A$ listener,located at $L$,directly in front of $S_{2}$,finds that the intensity is at a minimum when he is $2\, m$ away from $S_{2}$. The listener moves away from $S_{1}$,keeping his distance from $S_{2}$ fixed. The adjacent maximum of intensity is observed when the listener is at a distance $d$ from $S_{1}$. Then,$d$ is $......\, m$.

When two light waves each of amplitude $A$ and having a phase difference of $\frac{\pi}{2}$ are superimposed,the amplitude of the resultant wave is:

The equations of two waves are given as
$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}$
If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ 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.

Three harmonic waves having equal frequency $v$ and same intensity $I_{0}$,have phase angles $0, \frac{\pi}{4}$ and $-\frac{\pi}{4}$ respectively. When they are superimposed,the intensity of the resultant wave is close to

Two loudspeakers ($L_1$ and $L_2$) are placed with a separation of $10 \ m$, as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. $A$ voice recorder, initially at point $A$, at equidistance to both loudspeakers, is moved by $25 \ m$ along the line $AB$ while monitoring the audio signal. The measured signal was found to undergo $10$ cycles of minima and maxima during the movement. The frequency of the input signal is . . . . . . $Hz$. (Speed of sound in air is $324 \ m/s$ and $\sqrt{5} = 2.23$)