Question

AIIMS 2017
40.
Two sound waves having pressure
P, = 2 x 104 sin (21 x 104 t) Pa and
P, = 4 x 104 sin (31 x 104 + + 7/6) Pa
superimpose with each other. Find the amplitude of
resultant wave.
(1) 4.47 x 104 Pa
(2) 4.47 Pa
(3) 5.67 x 104 Pa
(4) 5.67 Pa​

Answer 1

Answer:

Resultant amplitude of two waves is given as

$R = 5.67 \times 10^4 Pa$

Explanation:

As we know that the two sound waves are given as

[tex]P_1 = 2\times 10^4 sin ( 2\pi  \times 10^4 t) Pa[/tex]

[tex]P_2 = 4 \times 10^4 sin( 3\pi \times 10^4 t + \frac{\pi}{6}) Pa[/tex]

Now we will have

[rex]R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 cos\theta}[/tex]

so we have

$R = \sqrt{(2 \times 10^4)^2 + (4 \times 10^4)^2 + 2(2\times 10^4)(4 \times 10^4) cos\frac{\pi}{6}}$

$R = 5.67 \times 10^4 Pa$

#Learn

Topic : superposition of waves

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