The equations of two waves are given as
$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}$
If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ is
$\begin{aligned}
& y_1=a \sin \left(\omega t+\phi_1\right) \\
& y_2=a \sin \left(\omega t+\phi_2\right)
\end{aligned}$
If the amplitude and time period of the resultant wave are the same as those of the individual waves,then $(\phi_1-\phi_2)$ is
- A$\cos ^{-1}\left(-\frac{1}{2}\right)$
- B$\cos ^{-1}\left(-\frac{1}{4}\right)$
- C$\cos ^{-1}\left(-\frac{1}{6}\right)$
- D$\cos ^{-1}\left(-\frac{1}{8}\right)$
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Two waves are simultaneously passing through a string and their equations are:
${y}_{1} = {A}_{1} \sin {k}({x} - {vt}), {y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0}).$
Given amplitudes ${A}_{1} = 12 \, {mm}$ and ${A}_{2} = 5 \, {mm}$,${x}_{0} = 3.5 \, {cm}$,and wave number ${k} = 6.28 \, {cm}^{-1}$. The amplitude of the resulting wave will be $...... \, {mm}$.
${y}_{1} = {A}_{1} \sin {k}({x} - {vt}), {y}_{2} = {A}_{2} \sin {k}({x} - {vt} + {x}_{0}).$
Given amplitudes ${A}_{1} = 12 \, {mm}$ and ${A}_{2} = 5 \, {mm}$,${x}_{0} = 3.5 \, {cm}$,and wave number ${k} = 6.28 \, {cm}^{-1}$. The amplitude of the resulting wave will be $...... \, {mm}$.
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View SolutionTwo identical sinusoidal waves are moving in the same direction along a stretched string and interfere with each other. The phase difference between them is $120^{\circ}$. The amplitudes of both the waves are the same. If the amplitude of the resultant wave due to interference is $2 \,mm$, the amplitude of each wave is:
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