Physics - Sound Wave Question with Solution | TestHub

PhysicsSound WaveInterferenceHard2 minPYQ_2020

PhysicsHardnumerical

The equations of three waves are given by $y_{1} = A_{0} \sin (k x - ω t)$ , $y_{2} = 3 \sqrt{2} A_{0} \sin (k x - ω t + ϕ)$ and $y_{3} = 4 A_{0} \cos (k x - ω t)$ . Theses waves are in the same direction and are superimposed. The phase difference between the resultant-wave and the first wave is $\frac{π}{4}$ and $ϕ = \frac{π}{n} \leq \frac{π}{2}$ , then what is the value of $n$ ?

Answer:

12.00

Solution:

$\tan (\frac{π}{4}) = \frac{BC}{AC} = \frac{A_{0} (4 + 3 \sqrt{2} \sin ϕ)}{A_{0} (1 + 3 \sqrt{2} \cos ϕ)}$

$\Rightarrow \cos ϕ - \sin ϕ = \frac{1}{\sqrt{2}}$
Squaring both sides, $\Rightarrow \cos^{2} ϕ + \sin^{2} ϕ - 2 \cos ϕ \sin ϕ = \frac{1}{2}$
$\Rightarrow 2 \sin ϕ \cos ϕ = \frac{1}{2}$
$\Rightarrow \sin 2 ϕ = \frac{1}{2}$
$\Rightarrow ϕ = \frac{1}{2} \sin^{- 1} (\frac{1}{2}) = \frac{π}{12}$

Stream:JEESubject:PhysicsTopic:Sound WaveSubtopic:Interference

⏱ 2mℹ️ Source: PYQ_2020

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