Two loudspeakers $L_1$ and $L_2$,driven by a common oscillator and amplifier,are arranged as shown. The frequency of the oscillator is gradually increased from zero,and the detector at $D$ records a series of maxima and minima. If the speed of sound is $330 \, m/s$,then the frequency at which the first maximum is observed is .... $Hz$.

Two waves have the same amplitude $A$ and the same frequency $\omega$. If the phase difference between them is $\pi / 2$,what are the resultant amplitude and the resultant frequency when they superimpose at a point?

The sources of sound $A$ and $B$ produce a wave of $350 Hz$ in the same phase. $A$ particle $P$ is vibrating under the influence of these two waves. If the amplitudes at $P$ produced by the two waves are $0.3 mm$ and $0.4 mm$,the resultant amplitude of the point $P$ will be,when $AP-BP=25 cm$ and the velocity of sound is $350 m s^{-1}$. (in $mm$)

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 $\sqrt{5} = 2.23$)

The equation of a simple harmonic progressive wave is given by $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$. The resultant amplitude of the wave is:

Two waves with same amplitude and frequency superpose at a point. The ratio of resultant intensities when they arrive in phase to that when they arrive $90^{\circ}$ out of phase is $\left[\cos \frac{\pi}{2}=0\right]$