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

Fundamental frequency of a closed organ pipe:

where l$l$ is the length of the resonating air column.

Core Logic

Initially, the full air column length is l$l$. Filling (1)/(5)$\frac{1}{5}$ of its space with fluid reduces the available vibrating air column length to:

Step 1: Calculate New Frequency

The modified acoustic frequency response is:

Step 2: Determine Percentage Shift

Pattern Recognition

Shortening the resonance tube length raises pitch frequency inversely (f ∝ (1)/(l)$f \propto \frac{1}{l}$). Reducing the air column length to 80%$80\%$ of its original value drives the frequency up to 125%$125\%$, yielding a positive 25%$25\%$ increase.

Evaluation Rubric / Model Answer

Option A: 25%

Chapter Mix

Class 11 Physics: Waves

Related Formula

Δ xmin = (λ)/(2)$$\Delta x_{\text{min}} = \frac{\lambda}{2}$$ [cite: 715]

Core Logic

From the given wave equation, the wave number k$k$ is the coefficient of x$x$ [cite: 120]:

Now find the wavelength λ$\lambda$: [cite: 717]

λ = (2π)/(k) = (2π)/(20π) = (1)/(10) m = 10 cm$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{20\pi} = \frac{1}{10}\ \text{m} = 10\ \text{cm}$$ [cite: 717]

The minimum distance between any two points moving with identical speeds in a continuous wave cycle corresponds to a phase difference of π$\pi$, which translates spatially to a half-wavelength separation ((λ)/(2)$\frac{\lambda}{2}$) [cite: 715]:

Distance = (λ)/(2) = (10)/(2) = 5 cm$$\text{Distance} = \frac{\lambda}{2} = \frac{10}{2} = 5\ \text{cm}$$ [cite: 717]

Pattern Recognition

Oscillating speed configuration reaches identical value fields twice per spatial wavelength cycle[cite: 715]. Thus, minimum distance separation scales exactly to (λ)/(2)$\frac{\lambda}{2}$[cite: 715].

Chapter Mix

Class 11 Physics: Waves

Related Formula

Fundamental frequency of a sonometer wire:

Beat frequency equation:

Core Logic

Let the tuning fork frequency be f$f$. Initially, it resonates with the wire at 6~N$6\mathrm{~N}$:

When the tension shifts to 54~N$54\mathrm{~N}$, the wire's new frequency becomes:

Taking the ratio of frequencies:

Step 1: Apply Beat Conditions

The new frequency f₂$f_2$ creates 12 beats per second with the fork:

Pattern Recognition

Resonance establishes a direct anchor equality (ffork = finitial$f_{\text{fork}} = f_{\text{initial}}$). Since tension changes by a factor of 9 (54/6$54/6$), the frequency scales up by a factor of 3 (√(9)$\sqrt{9}$).

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Oscillations

Related Formula

Core Logic

For a point source, the intensity I$I$ is inversely proportional to the square of the distance from the source. If the given intensity is at r₀ = 1m$r_0 = 1\mathrm{m}$ (usually implied if "at origin" is stated alongside a base value, though the phrasing "at the origin" is ambiguous), we can assume I₀ = 16 × 10⁻⁸$I_0 = 16 \times 10^{-8}$. The official NTA answer assumes the initial given intensity was at r=2m$r=2\mathrm{m}$. Wait, if I = 16 × 10⁻⁸$I = 16 \times 10^{-8}$ was the source power factor or the intensity at r=2$r=2$, let's reverse-engineer the answer `3`. If I ∝ 1/r²$I \propto 1/r^2$, and I₁$I_1$ at r=2$r=2$ and I₂$I_2$ at r=4$r=4$: I₂ = I₁ (r₁/r₂)² = I₁ (2/4)² = I₁ / 4$I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$. Difference Δ I = I₁ - I₂ = I₁ - I₁/4 = 3 I₁ / 4$\Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$. If this difference equals 3, then I₁$I_1$ must be 4$4$. But 16 × 10⁻⁸$16 \times 10^{-8}$ is given. Wait, if I$I$ at 1m$1\mathrm{m}$ is 16 × 10⁻⁸$16 \times 10^{-8}$: I(2m) = (16)/(2²) = 4 × 10⁻⁸$I(2\mathrm{m}) = \frac{16}{2^2} = 4 \times 10^{-8}$. I(4m) = (16)/(4²) = 1 × 10⁻⁸$I(4\mathrm{m}) = \frac{16}{4^2} = 1 \times 10^{-8}$. Δ I = 4 - 1 = 3 × 10⁻⁸$\Delta I = 4 - 1 = 3 \times 10^{-8}$. This perfectly matches.

Step 1: Calculate Intensity at r = 2m and r = 4m

Assume the intensity I₀ = 16 × 10⁻⁸ ~W m⁻²$I_0 = 16 \times 10^{-8} \mathrm{~W m}^{-2}$ represents the reference intensity at r=1m$r=1\mathrm{m}$. Intensity at r=2 ~m$r=2 \mathrm{~m}$:

Intensity at r=4 ~m$r=4 \mathrm{~m}$:

Step 2: Calculate the Difference

Magnitude of intensity difference:

Pattern Recognition

Although the question text ("at the origin") was poorly drafted (intensity at r=0$r=0$ would be infinite), the numerical setup expects you to treat 16 × 10⁻⁸$16 \times 10^{-8}$ as the P/4π$P/4\pi$ constant multiplier for the 1/r²$1/r^2$ dropoff BY NTA 3 BY RANKBIT (BONUS) Question is wrong as data is incomplete.

Chapter Mix

Class 11 Physics: Waves

Related Formula

Core Logic

Adding water to a closed organ pipe reduces its effective air column length, thereby increasing the fundamental frequency. The change in length corresponds directly to the volume of water poured.

Step 1: Calculate Initial Length

Step 2: Calculate Final Length

Step 3: Calculate Water Mass

The change in length (water depth) is:

Change in volume (volume of water) = Area × Δ$\times \Delta \ell$:

Mass of water (M$M$):

Pattern Recognition

Always map frequency changes to length changes using f ∝ 1/L$f \propto 1/L$. Poured water height is exactly the length change Δ L$\Delta L$. Watch out for CGS unit conversion (cm, grams).

Chapter Mix

Class 11 Physics: Waves