Two generators $S_1$ and $S_2$ produce water waves of equal frequency. $A$ point $P$ is located such that $(S_1P - S_2P)$ is equal to half a wavelength. When operated alone,$S_1$ produces an oscillation of amplitude $2a$ at $P$,while $S_2$ produces an oscillation of amplitude $a$. If the generators are operated in phase,which graph correctly shows the resultant oscillation at $P$?

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 $A_1 > A_2$]

The displacement of a particle executing periodic motion is given by :
$y = 4 \cos^2(t/2) \sin(1000t)$
This expression may be considered to be a result of the superposition of

Two small identical speakers are connected in phase to the same source. The speakers are $3 \,m$ apart and at ear level. An observer stands at $P, 4 \,m$ in front of one speaker as shown. The sound she hears is least intense when the wavelength is $\lambda_1$ and most intense when the wavelength is $\lambda_2$. Then,the possible values of $\lambda_1$ and $\lambda_2$ are

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 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 $......\, m$.