MCQ
Three waves of equal frequency having amplitudes $10\mu m,$ $4\mu m,$ $7\mu m$ 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
- A$4$
- ✓$5$
- C$6$
- D$7$
✓
Answer
Correct option: B.
$5$
b
(b) The amplitudes of the waves are
$a1 = 10 \mu m, a2 = 4 \mu m \,\,and\,\, a3 = 7\mu m$
and the phase difference between $1^{st}$ and $2^{nd}$ wave is $\frac{\pi }{2}$ and that between $2^{nd}$ and $3^{rd}$ wave is $\frac{\pi }{2}.$ Therefore, phase difference between $1^{st}$ and $3^{rd}$ is $\pi$. Combining $1^{st}$ with $3^{rd}$, their resultant amplitude is given by
$A_1^2 = a_1^2 + a_3^2 + 2{a_1}{a_3}\cos \phi $
or ${A_1} = \sqrt {{{10}^2} + {7^2} + 2 \times 10 \times 7\,\cos \pi } = \sqrt {100 + 49 - 140} $
$ = \sqrt 9 = 3\mu m$ in the direction of first.
Now combining this with $2^{nd}$ wave we have, the resultant amplitude
${A^2} = A_1^2 + a_2^2 + 2{A_1}{a_2}\cos \frac{\pi }{2}$
or $A = \sqrt {{3^2} + {4^2} + 2 \times 3 \times 4\cos {{90}^o}} = \sqrt {9 + 16} = 5\mu m$
(b) The amplitudes of the waves are
$a1 = 10 \mu m, a2 = 4 \mu m \,\,and\,\, a3 = 7\mu m$
and the phase difference between $1^{st}$ and $2^{nd}$ wave is $\frac{\pi }{2}$ and that between $2^{nd}$ and $3^{rd}$ wave is $\frac{\pi }{2}.$ Therefore, phase difference between $1^{st}$ and $3^{rd}$ is $\pi$. Combining $1^{st}$ with $3^{rd}$, their resultant amplitude is given by
$A_1^2 = a_1^2 + a_3^2 + 2{a_1}{a_3}\cos \phi $
or ${A_1} = \sqrt {{{10}^2} + {7^2} + 2 \times 10 \times 7\,\cos \pi } = \sqrt {100 + 49 - 140} $
$ = \sqrt 9 = 3\mu m$ in the direction of first.
Now combining this with $2^{nd}$ wave we have, the resultant amplitude
${A^2} = A_1^2 + a_2^2 + 2{A_1}{a_2}\cos \frac{\pi }{2}$
or $A = \sqrt {{3^2} + {4^2} + 2 \times 3 \times 4\cos {{90}^o}} = \sqrt {9 + 16} = 5\mu m$
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