Two periodic waves of amplitudes A1 and A2 pass thorough a region- If A1-A2- the difference in the maximum and minimum resultant amplitude possible isa 2 A1b 2 A2c A1-A2d A1-A2

(b) 2A2 We know resultant amplitude is given by Anet=A12+A22+2A1A2cos #981; For maximum resultant amplitude Amax=A1+A2 For minimum resultant amplitude Amin=A ...

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Standard X

Physics

Resonance in Sound Waves

Two periodic ...

Question

Two periodic waves of amplitudes A₁ and A₂ pass thorough a region. If A₁ > A₂, the difference in the maximum and minimum resultant amplitude possible is
(a) 2A₁
(b) 2A₂
(c) A₁ + A₂
(d) A₁ − A₂.

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Solution

(b) 2A₂

We know resultant amplitude is given by
$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
For maximum resultant amplitude
$A_{\max} = A_{1} + A_{2}$
For minimum resultant amplitude
$A_{\min} = A_{1} - A_{2}$

So, the difference between A_max and A_min is
$A_{\max} - A_{\min} = A_{1} + A_{2} - A_{1} + A_{2} = 2 A_{2}$

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Two periodic waves of amplitudes A1 and A2 pass thorough a region. If A1 > A2, the difference in the maximum and minimum resultant amplitude possible is (a) 2A1 (b) 2A2 (c) A1 + A2 (d) A1 − A2.

(b) 2A2 We know resultant amplitude is given by Anet=A12+A22+2A1A2cosϕ For maximum resultant amplitude Amax=A1+A2 For minimum resultant amplitude Amin=A1-A2 So, the difference between Amax and Amin is Amax-Amin=A1+A2-A1+A2=2A2

Two periodic waves of amplitudes A₁ and A₂ pass thorough a region. If A₁ > A₂, the difference in the maximum and minimum resultant amplitude possible is
(a) 2A₁
(b) 2A₂
(c) A₁ + A₂
(d) A₁ − A₂.