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Two waves ${{y}_{1}}={{A}_{1}}\sin (\omega t-{{\beta }_{1}})$, ${{y}_{2}}={{A}_{2}}\sin (\omega t-{{\beta }_{2}})$ Superimpose to form a resultant wave whose amplitude is
If the ratio of amplitude of wave is 2 : 1, then the ratio of maximum and minimum intensity is
The two interfering waves have intensities in the ratio 9 : 4. The ratio of intensities of maxima and minima in the interference pattern will be
If the ratio of amplitude of two waves is 4 : 3. Then the ratio of maximum and minimum intensity will be
Equation of motion in the same direction is given by ${{y}_{1}}=A\sin (\omega t-kx)$, ${{y}_{2}}=A\sin (\omega t-kx-\theta )$. The amplitude of the medium particle will be
Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to the minimum intensity, is equal to
The displacement of the interfering light waves are ${{y}_{1}}=4\sin \omega \,t$ and ${{y}_{2}}=3\sin \left( \omega \,t+\frac{\pi }{2} \right)$. What is the amplitude of the resultant wave
Two waves are represented by ${{y}_{1}}=a\sin \left( \omega \,t+\frac{\pi }{6} \right)$ and ${{y}_{2}}=a\cos \omega \,t$. What will be their resultant amplitude
The amplitude of a wave represented by displacement equation $y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\cos \omega t$ will be
Two waves having equations ${{x}_{1}}=a\sin (\omega \,t+{{\varphi }_{1}})$, ${{x}_{2}}=a\sin \,(\omega \,t+{{\varphi }_{2}})$. If in the resultant wave the frequency and amplitude remain equal to those of superimposing waves. Then phase difference between them is

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