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 $A_1 > A_2$]

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:

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 are represented by $y_1 = a \sin(\omega t + \frac{\pi}{6})$ and $y_2 = a \cos(\omega t)$. What will be their resultant amplitude?

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