When two sound waves having amplitude $3$ and $5$ units are superimposed,then the ratio of maximum to minimum intensity of the wave produced is:

Two simple harmonic waves having equal amplitudes of $8\,cm$ and equal frequency of $10\,Hz$ are moving along the same direction. The resultant amplitude is also $8\,cm$. The phase difference between the individual waves is $..................$ degree.

Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0, \pi / 3, 2 \pi / 3$ and $\pi$. When they are superposed,the intensity of the resulting wave is $nI_0$. The value of $n$ is

If two waves represented by $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{3})$ interfere at a point,then the amplitude of the resulting wave will be about:

Two sound waves (expressed in $CGS$ units) given by $y_1 = 0.3 \sin \frac{2\pi}{\lambda}(vt - x)$ and $y_2 = 0.4 \sin \frac{2\pi}{\lambda}(vt - x + \theta)$ interfere. The resultant amplitude at a place where phase difference is $\pi/2$ will be .... $cm$.

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