Answer
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Hint: We will write the magnitude of amplitude of the three waves by using the formula for the sum of amplitudes. First, we will add two waves and then we will add the third wave to their resultant. We can add them in any order if we keep the phase difference consistent.
Formula used:
Resultant amplitude of sum of two waves
${{A}^{2}}={{A}_{1}}^{2}+{{A}_{2}}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi $
Complete answer:
The waves have the same frequency so their amplitudes can directly be calculated using the formula given. First, we will add up the two waves with an amplitude of 10$\mu m$ and 7$\mu m$. They both will have a phase difference equal to twice of $\dfrac{\pi }{2}$ i.e. $\pi $. Hence the resultant amplitude will be given as
$\begin{align}
& {{A}^{2}}={{10}^{2}}+{{7}^{2}}+2\times 10\times 7\times \cos \left( {{180}^{o}} \right)={{10}^{2}}+{{7}^{2}}-2\times 10\times 7=100+49-140=9 \\
& A=3\mu m \\
\end{align}$
This wave will be in the same phase as the 10$\mu m$ wave as here destructive interference takes place and the phase of the wave with the larger amplitude will be maintained. Now, we will add the remaining wave of 4$\mu m$ with this. The phase difference here will be $\dfrac{\pi }{2}$. Hence the resultant amplitude will be given as
$\begin{align}
& {{A}^{2}}={{3}^{2}}+{{4}^{2}}+2\times 3\times 4\times \cos \left( {{90}^{o}} \right)={{3}^{2}}+{{4}^{2}}=9+16=25 \\
& A=5\mu m \\
\end{align}$
The resultant amplitude of the three waves will be 5$\mu m$.
Hence, the correct option is C, i.e. 5.
Note:
Students must take care that although we could take the sum of any two waves first, we must take the sum of the first and the third wave first for simplicity. If we take the sum of the first and the second wave first, the phase difference of the resultant of the third wave will have to be calculated and that would make the problem more complex. Simply adding the first and the third wave first we can skip this problem and get the answer easily.
Formula used:
Resultant amplitude of sum of two waves
${{A}^{2}}={{A}_{1}}^{2}+{{A}_{2}}^{2}+2{{A}_{1}}{{A}_{2}}\cos \phi $
Complete answer:
The waves have the same frequency so their amplitudes can directly be calculated using the formula given. First, we will add up the two waves with an amplitude of 10$\mu m$ and 7$\mu m$. They both will have a phase difference equal to twice of $\dfrac{\pi }{2}$ i.e. $\pi $. Hence the resultant amplitude will be given as
$\begin{align}
& {{A}^{2}}={{10}^{2}}+{{7}^{2}}+2\times 10\times 7\times \cos \left( {{180}^{o}} \right)={{10}^{2}}+{{7}^{2}}-2\times 10\times 7=100+49-140=9 \\
& A=3\mu m \\
\end{align}$
This wave will be in the same phase as the 10$\mu m$ wave as here destructive interference takes place and the phase of the wave with the larger amplitude will be maintained. Now, we will add the remaining wave of 4$\mu m$ with this. The phase difference here will be $\dfrac{\pi }{2}$. Hence the resultant amplitude will be given as
$\begin{align}
& {{A}^{2}}={{3}^{2}}+{{4}^{2}}+2\times 3\times 4\times \cos \left( {{90}^{o}} \right)={{3}^{2}}+{{4}^{2}}=9+16=25 \\
& A=5\mu m \\
\end{align}$
The resultant amplitude of the three waves will be 5$\mu m$.
Hence, the correct option is C, i.e. 5.
Note:
Students must take care that although we could take the sum of any two waves first, we must take the sum of the first and the third wave first for simplicity. If we take the sum of the first and the second wave first, the phase difference of the resultant of the third wave will have to be calculated and that would make the problem more complex. Simply adding the first and the third wave first we can skip this problem and get the answer easily.
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