Three waves of equal frequency having amplitudes $10 \,\mu m, 4 \,\mu m$ and $7 \,\mu m$ arrive at a given point with successive phase difference of $\frac{\pi}{2}$. The amplitude of the resulting wave in $\mu m$ is given by

The amplitude of a wave,represented by the displacement equation $y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$,will be:

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

Two waves represented by $y_1 = a \sin \frac{2\pi}{\lambda} (vt - x)$ and $y_2 = a \cos \frac{2\pi}{\lambda} (vt - x)$ are superposed. The resultant wave has an amplitude equal 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 $......\, m$.

The resultant amplitude of waves ${y_1} = a \sin \left( \omega t + \frac{\pi}{3} \right)$ and ${y_2} = a \sin \omega t$ is: