Two loudspeakers L₁$L_{1}$ and L₂$L_{2}$ are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point A, at equidistance to both loud speakers, is moved by 25 m along the line AB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is \_\_\_\_ Hz (Speed of sound in air is 324~m/s$324~\mathrm{m/s}$ and √(5)=2.23$\sqrt{5}=2.23$)

Related Formula

where, A₁, A₂$A_1, A_2$ = amplitudes of individual harmonic waves φ$\phi$ = phase difference between the waves Aᵣₑₛ$A_{\text{res}}$ = resultant amplitude θ$\theta$ = resultant phase angle relative to the first wave

Core Logic

Given parameters:

• A₁ = 4$A_1 = 4$
• A₂ = 2$A_2 = 2$
• Phase difference, φ = (2π)/(3) = 120^°$\phi = \frac{2\pi}{3} = 120^\circ$



Calculate Resultant Amplitude:

Calculate Resultant Phase (angle θ$\theta$):

Step 1: Result Format

Represent the resultant amplitude and phase as a pair:

Pattern Recognition

Sees: Vector-like addition of wave amplitudes. Shortcut: Solve it like a vector addition problem. Vector A₁$A_1$ along horizontal (0$0$), and Vector A₂$A_2$ at 120^°$120^\circ$. The magnitude is √(4²+2²-2(4)(2)(0.5)) = 2√(3)$\sqrt{4^2+2^2-2(4)(2)(0.5)} = 2\sqrt{3}$ and direction is π/6$\pi/6$. ✓

Chapter Mix

Class 11 Physics: Waves

Related Formula

where, v$v$ = velocity of transverse wave on a string T$T$ = tension on string μ$\mu$ = linear mass density ρ$\rho$ = density of material A$A$ = cross-sectional area

Core Logic

Since both strings are made of the same material, their density ρ$\rho$ is the same. Also, they are stretched to have the same amount of tension T$T$.

Substitute the formula for μ$\mu$ into the velocity equation:

Thus, the wave velocity is inversely proportional to the radius of the cross section of the string:

Step 1: Ratio Computation

Given:

• R₁ = R$R_1 = R$
• R₂ = R/2$R_2 = R/2$

Substitute these values:

Pattern Recognition

Sees: Circular cross-section strings + transverse wave speed relation. Shortcut: Wave velocity v ∝ 1√(μ) ∝ (1)/(R)$v \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{R}$. If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by √(4) = 2$\sqrt{4} = 2$ times. ✓

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

Assertion A: Sound velocity relies on structural elasticity bounds. Solids are highly rigid compared to fluids, making speed significantly higher. (True)

Reason R: Solids resist structural compression far better than unbonded gases, giving them significantly higher Bulk Modulus properties. Thus, statement R is completely false.

Step 1: Final Conclusion

Assertion A is true, but Reason R is false, aligning perfectly with option (4).

Pattern Recognition

Even though density ρ$\rho$ is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

From the given wave equation, angular frequency ω = 628 rad/s$\omega = 628\text{ rad/s}$. Given wave velocity v = 300 m/s$v = 300\text{ m/s}$. Using the relation v = (ω)/(K)$v = \frac{\omega}{K}$:

Step 1: Compute Wavelength

Substitute K = (2π)/(λ)$K = \frac{2\pi}{\lambda}$:

Since 2π ≈ 2 × 3.14 = 6.28$2\pi \approx 2 \times 3.14 = 6.28$, the expression simplifies neatly:

Pattern Recognition

Notice standard values like ω = 628 = 200π$\omega = 628 = 200\pi$, which means the frequency is exactly 100 Hz$100\text{ Hz}$. Using v = fλ 300 = 100λ λ = 3 m$v = f\lambda \implies 300 = 100\lambda \implies \lambda = 3\text{ m}$ avoids setting up fractions.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Laplace correction equation for speed of sound:

Given that (P)/(ρ)$\frac{P}{\rho}$ is constant for all three gases:

where γ = 1 + (2)/(f)$\gamma = 1 + \frac{2}{f}$ (adiabatic constant).

Core Logic

Determine the γ$\gamma$ factor based on molecular atomic structures:

• He$\mathrm{He}$ (Monatomic) f = 3 γHe = (5)/(3)$\implies f = 3 \implies \gamma_{\mathrm{He}} = \frac{5}{3}$
• CH₄$\mathrm{CH}_{4}$ (Polyatomic/Non-linear) γ_CH₄ ≈ (4)/(3)$\implies \gamma_{\mathrm{CH}_{4}} \approx \frac{4}{3}$ based on experimental references.
• CO₂$\mathrm{CO}_{2}$ (Triatomic linear/vibrational modes) γ_CO₂ ≈ (4)/(3)$\implies \gamma_{\mathrm{CO}_{2}} \approx \frac{4}{3}$ as provided in textbook standard testing matrices.

Step 1: Construct the Ratio

Substitute these values into the proportionality:

Pattern Recognition

When (P)/(ρ)$\frac{P}{\rho}$ is locked down constant, sound speed depends strictly on internal degrees of freedom via γ$\gamma$. Keep standard experimental values of complex gases like CH₄$\mathrm{CH}_4$ and CO₂$\mathrm{CO}_2$ memorized.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory