6. (JEE Main 2021 (Online) 26th August Evening Shift)

Two waves are simultaneously passing through a string and their equations are :

y₁ = A₁ sin k(x $-$ vt), y₂ = A₂ sin k(x $-$ vt + x₀). Given amplitudes A₁ = 12 mm and A₂ = 5 mm, x₀ = 3.5 cm and wave number k = 6.28 cm^{$-$ 1}. The amplitude of resulting wave will be ................ mm.

Correct Answer is 7

y₁ = A₁ sin k(x $-$ vt)

y₁ = 12 sin 6.28 (x $-$ vt)

y₂ = 5 sin 6.28 (x $-$ vt + 3.5)

$Δ ϕ = \frac{2 π}{λ} (Δ x)$

$= K (Δ x)$

$= 6.28 \times 3.5 = \frac{7}{2} \times 2 π = 7 π$

$A_{n e t} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$

$A_{n e t} = \sqrt{{(12)}^{2} + {(5)}^{2} + 2 (12) (5) \cos (7 π)}$

$= \sqrt{144 + 25 - 120}$ F

7. (JEE Main 2020 (Online) 9th January Morning Slot)

Three harmonic waves having equal frequency $ν$ and same intensity $I_{0}$ , have phase angles 0, $\frac{π}{4}$ and $- \frac{π}{4}$ respectively. When they are superimposed the intensity of the resultant wave is close to :

A. 5.8 I₀

B. 3 I₀

C. 0.2 I₀

D. I₀

Correct Answer is Option (A)

JEE Main 2020 (Online) 9th January Morning Slot Physics - Waves Question 64 English Explanation

I₀ = CA²

A_R = A + A $\sqrt{2}$ = A(1 + $\sqrt{2}$ )

I_R = C $A_{R}^{2}$

$∴$ I_R = CA² ${(\sqrt{2} + 1)}^{2}$ = 5.8 I₀

8. (JEE Main 2019 (Online) 12th April Evening Slot)

A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound ? [Given reference intensity of sound as 10^–12 W/m² ]

A. 20 cm

B. 10 cm

C. 30 cm

D. 40 cm

Correct Answer is Option (C)

Sound level = 10 $\log_{10} (\frac{I}{I_{0}})$

$\Rightarrow$ 120 = 10 $\log_{10} (\frac{I}{10^{- 12}})$

$\Rightarrow$ 12 = $\log_{10} (\frac{I}{10^{- 12}})$

$\Rightarrow$ 10¹² = $\frac{I}{10^{- 12}}$

$\Rightarrow$ I = 1 W/m²

Also we know,

I = $\frac{P}{4 π r^{2}}$

$\Rightarrow$ 1 = $\frac{2}{4 π r^{2}}$

$\Rightarrow$ r = 40 cm

9. (JEE Main 2019 (Online) 10th April Evening Slot)

The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz, is :

A. JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 75 English Option 1

B. JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 75 English Option 2

C. JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 75 English Option 3

D. JEE Main 2019 (Online) 10th April Evening Slot Physics - Waves Question 75 English Option 4

Correct Answer is Option (C)

Beat frequency = |f₁ – f₂| = 11 – 9 = 2 Hz

10. (AIEEE 2007)

A sound absorber attenuates the sound level by $20$ $d B$ . The intensity decreases by a factor of

A. 100

B. 1000

C. 10000

D. 10

Correct Answer is Option (A)

We have, $L_{1} = 10 \log (\frac{I_{1}}{I_{0}});$

$L_{2} = 10 \log (\frac{I_{2}}{I_{0}})$

$∴$ $L_{1} - L_{2} = 10 \log (\frac{I_{1}}{I_{0}}) - 10 \log (\frac{I_{2}}{I_{0}})$

or, $Δ L = 10 \log (\frac{I_{1}}{I_{0}} \times \frac{I_{0}}{I_{2}})$

or, $Δ L = 10 \log (\frac{I_{1}}{I_{2}})$

or, $20 = 10 \log (\frac{I_{1}}{I_{2}})$

or, $2 = \log (\frac{I_{1}}{I_{2}})$

or, $\frac{I_{1}}{I_{2}} = 10^{2}$

or, $I_{2} = \frac{I_{1}}{100} .$

$\Rightarrow$ Intensity decreases by a factor $100.$