Welcome to Lecture 125 (Master Index Entry: L119) of our comprehensive JEE Physics Master Series! Advancing through the foundational wave acoustic concepts of Unit 10: Oscillations and Waves, we present the definitive analytical masterclass: Beats in Sound Waves — Physical Mechanism of Formation, Rigorous Mathematical Derivation, Time Intervals of Waxing and Waning, Tuning Fork Loading & Filing Dynamics, and Doppler-Coupled Beat Systems. When two sound waves possessing slightly different frequencies propagate simultaneously through the same medium in the same direction, their superposition produces a rhythmic, periodic rise and fall in acoustic intensity known as beats. In this masterclass, we construct the complete mathematical envelope theory, derive the exact expressions for waxing and waning, analyze tuning fork frequency shifts under mass loading and tip filing, examine multi-frequency superpositions, and deconstruct high-yield JEE Main and Advanced problem archetypes.
Detailed Concept Card: Physical & Mathematical Foundations of Beats
1. The Phenomenon of Beats & Persistence of Hearing
When two longitudinal sound waves of identical amplitude $A$ and slightly different frequencies $f_1$ and $f_2$ travel in the same direction, interference occurs in the time domain at any fixed point in space. The resultant acoustic intensity periodically waxes (reaches maximum loudness) and wanes (decreases to minimum intensity or silence). This acoustic modulation is termed beats.
Persistence of Hearing Constraint: The human ear retains a sensory impression of sound for approximately $\tau \approx 0.1\text{ s}$ ($1/10\text{th}$ of a second). To perceive distinct, periodic fluctuations in loudness, the time interval between consecutive beats must exceed this auditory retention threshold:
$$T_b = \frac{1}{|f_1 – f_2|} \ge 0.1\text{ s} \implies \mathbf{|f_1 – f_2| \le 10\text{ Hz}}$$
If the frequency difference exceeds $10\text{ Hz}$, the human brain cannot resolve individual beats; instead, it perceives a continuous, rough, or harsh acoustic sensation, or a third subjective difference tone.
2. Rigorous Mathematical Derivation of Beat Frequency
Consider two harmonic acoustic displacement waves of equal amplitude $A$, propagating along the $+x$-direction with slightly different angular frequencies $\omega_1 = 2\pi f_1$ and $\omega_2 = 2\pi f_2$, and corresponding wave numbers $k_1 = \omega_1/v$ and $k_2 = \omega_2/v$:
$$s_1(x, t) = A \sin(k_1 x – \omega_1 t)$$
$$s_2(x, t) = A \sin(k_2 x – \omega_2 t)$$
Without loss of generality, let us monitor the acoustic state at the spatial origin $x = 0$:
$$s_1(t) = A \sin(-\omega_1 t) = -A \sin(\omega_1 t)$$
$$s_2(t) = A \sin(-\omega_2 t) = -A \sin(\omega_2 t)$$
By the Principle of Superposition, the resultant displacement $s(t) = s_1(t) + s_2(t)$ is:
$$s(t) = -A [\sin(\omega_1 t) + \sin(\omega_2 t)]$$
Applying the standard prosthetic trigonometric identity $\sin C + \sin D = 2 \cos\left(\frac{C – D}{2}\right) \sin\left(\frac{C + D}{2}\right)$:
$$s(t) = -2A \cos\left(\frac{\omega_1 – \omega_2}{2} t\right) \sin\left(\frac{\omega_1 + \omega_2}{2} t\right)$$
Rewriting in terms of ordinary cyclic frequencies $f_1$ and $f_2$ ($\omega = 2\pi f$):
$$s(t) = \mathbf{A_{\text{mod}}(t) \sin(2\pi f_{\text{avg}} t)}$$
Where:
- Carrier Wave Frequency: $f_{\text{avg}} = \frac{f_1 + f_2}{2}$ (the high-frequency oscillation of air molecules).
- Time-Dependent Modulated Amplitude (Envelope): $$A_{\text{mod}}(t) = 2A \cos\left[\pi (f_1 – f_2) t\right]$$
3. Waxing, Waning & Beat Period Analysis
Because the acoustic intensity $I$ perceived by an observer is proportional to the square of the displacement amplitude ($I \propto A_{\text{mod}}^2$):
$$I(t) \propto 4A^2 \cos^2\left[\pi (f_1 – f_2) t\right] = 2A^2 \left\{1 + \cos\left[2\pi (f_1 – f_2) t\right]\right\}$$
| Acoustic Phase State | Mathematical Condition | Instantaneous Time ($t$) | Resultant Amplitude & Intensity | Temporal Separation ($\Delta t$) |
|---|---|---|---|---|
| Waxing (Intensity Maxima) | $\cos[\pi(f_1 – f_2)t] = \pm 1$ $\implies \pi(f_1 – f_2)t = m\pi$ |
$$t = \frac{m}{|f_1 – f_2|}$$ for $m = 0, 1, 2, 3 \dots$ |
$$A_{\max} = 2A$$ $$I_{\max} = 4A^2 \propto 4I_0$$ |
Period between consecutive maxima: $$T_b = \frac{1}{|f_1 – f_2|}$$ |
| Waning (Intensity Minima) | $\cos[\pi(f_1 – f_2)t] = 0$ $\implies \pi(f_1 – f_2)t = (m + \frac{1}{2})\pi$ |
$$t = \frac{2m + 1}{2|f_1 – f_2|}$$ for $m = 0, 1, 2, 3 \dots$ |
$$A_{\min} = 0$$ $$I_{\min} = 0\text{ (Silence)}$$ |
Period between consecutive minima: $$T_b’ = \frac{1}{|f_1 – f_2|}$$ |
Fundamental Beat Formulae:
- Beat Period ($T_b$): The time interval between two successive waxing events (or two successive waning events): $$\mathbf{T_b = \frac{1}{|f_1 – f_2|}}$$
- Beat Frequency ($f_b$): The number of intensity maxima (beats) heard per second: $$\mathbf{f_b = \frac{1}{T_b} = |f_1 – f_2|}$$
- Waxing-to-Waning Interval ($\Delta t_{\text{max}\to\text{min}}$): The time elapsed from a maximum to the immediately following minimum: $$\mathbf{\Delta t = \frac{T_b}{2} = \frac{1}{2|f_1 – f_2|}}$$
4. Beats with Unequal Wave Amplitudes ($A_1 \ne A_2$)
In real experimental conditions, the two superposing waves may possess unequal amplitudes $A_1$ and $A_2$. In this generalized regime:
- Maximum Resultant Amplitude: $A_{\max} = A_1 + A_2 \implies I_{\max} \propto (A_1 + A_2)^2$
- Minimum Resultant Amplitude: $A_{\min} = |A_1 – A_2| \implies I_{\min} \propto (A_1 – A_2)^2$
- Acoustic Contrast Ratio: $$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_1 + A_2}{A_1 – A_2}\right)^2$$ Physical Consequence: When $A_1 \ne A_2$, the waning phase is no longer complete silence ($I_{\min} > 0$). The beats are perceived as a modulation between loud and faint sound rather than complete extinction.
5. Tuning Fork Mechanics: Loading & Filing Dynamics
A tuning fork behaves as an elastic mechanical cantilever. Its fundamental natural vibration frequency is determined by prong elasticity (stiffness $k$) and effective inertial mass $m$:
$$f \propto \sqrt{\frac{k}{m}}$$
| Physical Modification | Inertial / Elastic Shift | Effect on Frequency ($f$) | Beat Frequency Evolution ($f_b = |f_A – f_B|$) |
|---|---|---|---|
| Loading Prongs with Wax | Mass increases ($m \uparrow$) while stiffness $k$ remains unchanged. | Frequency Decreases ($f \downarrow$) | • If $f_A > f_B$: $f_b$ decreases. • If $f_A < f_B$: $f_b$ increases. • If $f_A \gg f_B$ and heavily loaded: $f_b$ may decrease to 0, then reverse and increase. |
| Filing the Prongs (Tips) | Prong tip mass decreases ($m \downarrow$) while effective length reduces slightly. | Frequency Increases ($f \uparrow$) | • If $f_A < f_B$: $f_b$ decreases. • If $f_A > f_B$: $f_b$ increases. |
The Two-Branch Decision Protocol (Master JEE Algorithm):
When an unknown fork $A$ produces $b$ beats/sec with a standard fork $B$ of known frequency $f_B$, two mathematical possibilities exist:
$$f_A = f_B + b \quad \text{OR} \quad f_A = f_B – b$$
To eliminate ambiguity, a perturbing physical operation (loading or filing) is applied to fork $A$:
- If fork $A$ is loaded with wax ($f_A$ decreases to $f_A’$):
- If the new beat frequency $b’ < b$ (beats decrease), then $f_A$ was approaching $f_B$ from above $\implies \mathbf{f_A = f_B + b}$.
- If the new beat frequency $b’ > b$ (beats increase), then $f_A$ was moving further below $f_B \implies \mathbf{f_A = f_B – b}$.
- If the new beat frequency remains unchanged ($b’ = b$), then $f_A$ decreased by exactly $2b$, passing symmetrically through $f_B$ ($f_A – f_B = b$ and $f_B – f_A’ = b$) $\implies \mathbf{f_A = f_B + b}$.
- If fork $A$ is filed ($f_A$ increases to $f_A”$):
- If the new beat frequency $b” < b$ (beats decrease), then $f_A$ was approaching $f_B$ from below $\implies \mathbf{f_A = f_B - b}$.
- If the new beat frequency $b” > b$ (beats increase), then $f_A$ was moving further above $f_B \implies \mathbf{f_A = f_B + b}$.
6. Superposition of Multiple Frequencies ($N \ge 3$)
When three collinear acoustic waves of frequencies $f_1 < f_2 < f_3$ superpose simultaneously, pairwise interference produces three distinct beat components:
$$b_{12} = f_2 – f_1, \quad b_{23} = f_3 – f_2, \quad b_{13} = f_3 – f_1 = b_{12} + b_{23}$$
If the frequencies form an arithmetic progression (i.e., $f_1 = f_0 – \Delta f$, $f_2 = f_0$, $f_3 = f_0 + \Delta f$), the resultant sound intensity exhibits a fundamental periodic repetition rate equal to $\mathbf{\Delta f}$, with harmonic ripples at $2\Delta f$. The number of primary beats heard per second by the human ear is simply $\mathbf{\Delta f}$.
Four Graded Solved Examples
Example 1: Tuning Fork Inversion via Loading & Filing (Basic to Intermediate)
Problem Statement: A tuning fork $A$ produces $4\text{ beats/second}$ when sounded together with a standard tuning fork $B$ of frequency $f_B = 288\text{ Hz}$. When the prongs of fork $A$ are lightly loaded with wax, the beat frequency with fork $B$ reduces to $2\text{ beats/second}$.
(a) Determine the original frequency of tuning fork $A$.
(b) If instead of loading with wax, the prongs of the original fork $A$ were filed, calculate the resulting beat frequency with fork $B$, assuming filing increases its frequency by $3\text{ Hz}$.
Step-by-Step Solution:
1. Initial Ambiguity: The known fork has frequency $f_B = 288\text{ Hz}$. Since the initial beat frequency is $b = 4\text{ Hz}$, the possible frequencies for fork $A$ are:
$$f_A = f_B \pm 4\text{ Hz} \implies f_A = 292\text{ Hz} \quad \text{OR} \quad f_A = 284\text{ Hz}$$
2. Effect of Loading with Wax: Loading fork $A$ increases its effective mass, strictly lowering its frequency: $f_A’ < f_A$.
• Hypothesis 1 ($f_A = 292\text{ Hz}$): When $f_A$ decreases, its value moves closer to $288\text{ Hz}$, so $|f_A’ – 288| < 4$. The problem states that the beat frequency decreases to $2\text{ Hz}$ ($f_A' = 290\text{ Hz}$). This is physically consistent.
• Hypothesis 2 ($f_A = 284\text{ Hz}$): When $f_A$ decreases, its value moves further away from $288\text{ Hz}$ ($f_A’ < 284 \implies |f_A' - 288| > 4$). This contradicts the observed decrease in beats.
Therefore, the original frequency of fork $A$ is uniquely:
$$\mathbf{f_A = 292\text{ Hz}}$$
3. Part (b) — Filing Fork $A$: Filing the prongs of fork $A$ increases its frequency by $3\text{ Hz}$:
$$f_A” = f_A + 3\text{ Hz} = 292 + 3 = 295\text{ Hz}$$
The resulting beat frequency with fork $B$ ($288\text{ Hz}$) is:
$$\mathbf{f_b” = |f_A” – f_B| = |295 – 288| = 7\text{ beats/second}}$$
Example 2: Acoustic Pipe Thermal Shift Producing Beats (Intermediate Application)
Problem Statement: Two identical open organ pipes, each of physical length $L = 85.0\text{ cm}$, are tuned to resonate in their fundamental mode in unison at temperature $T_1 = 27^\circ\text{C}$ ($300\text{ K}$). The speed of sound in air at $27^\circ\text{C}$ is $v_1 = 340.0\text{ m/s}$. The temperature of the air in one pipe is raised to $T_2 = 39^\circ\text{C}$ ($312\text{ K}$) while the other pipe is maintained strictly at $27^\circ\text{C}$. Neglecting thermal expansion of the pipe material and end corrections, calculate the beat frequency produced when both pipes are sounded simultaneously in their fundamental modes.
Step-by-Step Solution:
1. Fundamental Frequency of First Pipe at $27^\circ\text{C}$ ($300\text{ K}$):
$$f_1 = \frac{v_1}{2L} = \frac{340.0}{2 \times 0.850} = \frac{340.0}{1.700} = \mathbf{200.0\text{ Hz}}$$
2. Speed of Sound in Second Pipe at $39^\circ\text{C}$ ($312\text{ K}$): Since sound velocity in an ideal gas scales as $v \propto \sqrt{T}$:
$$v_2 = v_1 \sqrt{\frac{T_2}{T_1}} = 340.0 \sqrt{\frac{312}{300}} = 340.0 \sqrt{1.04}$$
Using binomial approximation for small temperature increments ($\Delta T/T = 12/300 = 0.04$):
$$\sqrt{1 + 0.04} \approx 1 + \frac{1}{2}(0.04) = 1 + 0.02 = 1.02$$
$$v_2 \approx 340.0 \times 1.02 = 346.8\text{ m/s}$$
3. Fundamental Frequency of Second Pipe ($f_2$):
$$f_2 = \frac{v_2}{2L} = f_1 \sqrt{\frac{T_2}{T_1}} \approx 200.0 \times 1.02 = \mathbf{204.0\text{ Hz}}$$
4. Beat Frequency Calculation:
$$\mathbf{f_b = |f_2 – f_1| = 204.0\text{ Hz} – 200.0\text{ Hz} = 4.0\text{ beats/second}}$$
Exact evaluation check: $\sqrt{312/300} = \sqrt{1.04} = 1.0198039 \implies f_2 = 203.96\text{ Hz} \implies f_b = 3.96\text{ Hz} \approx 4\text{ Hz}$.
Example 3: Multi-Frequency Superposition & Waveform Synthesis (Advanced Theory)
Problem Statement: Three harmonic sound waves of equal amplitude $A$ and angular frequencies $\omega_1 = \omega_0 – \Delta\omega$, $\omega_2 = \omega_0$, and $\omega_3 = \omega_0 + \Delta\omega$ (where $\Delta\omega \ll \omega_0$) superpose at a point $x = 0$.
(a) Derive the analytical equation for the resultant displacement $s(t)$.
(b) Derive the time-dependent intensity expression $I(t)$ and find the ratio of maximum to minimum intensity ($I_{\max} / I_{\min}$).
(c) State the number of intensity peaks observed per unit time.
Step-by-Step Solution:
1. Superposition of the Three Waves:
$$s(t) = A \sin[(\omega_0 – \Delta\omega)t] + A \sin(\omega_0 t) + A \sin[(\omega_0 + \Delta\omega)t]$$
Combine the first and third terms using prosthaphaeresis:
$$A \sin[(\omega_0 – \Delta\omega)t] + A \sin[(\omega_0 + \Delta\omega)t] = 2A \cos(\Delta\omega t) \sin(\omega_0 t)$$
Factoring out the common carrier oscillation $\sin(\omega_0 t)$:
$$\mathbf{s(t) = \left[A + 2A \cos(\Delta\omega t)\right] \sin(\omega_0 t) = A\left[1 + 2\cos(\Delta\omega t)\right] \sin(\omega_0 t)}$$
2. Modulated Amplitude & Intensity: The amplitude envelope is:
$$A_{\text{mod}}(t) = A [1 + 2\cos(\Delta\omega t)]$$
The instantaneous intensity is proportional to $A_{\text{mod}}^2(t)$:
$$\mathbf{I(t) \propto A^2 \left[1 + 2\cos(\Delta\omega t)\right]^2}$$
3. Maximum & Minimum Intensity:
• Maximum Amplitude: Occurs when $\cos(\Delta\omega t) = +1$:
$$A_{\max} = A [1 + 2(1)] = 3A \implies \mathbf{I_{\max} \propto 9A^2}$$
• Minimum Amplitude: Occurs when $\cos(\Delta\omega t) = -1/2$:
$$A_{\min} = A [1 + 2(-1/2)] = 0 \implies \mathbf{I_{\min} = 0}$$
(Note: When $\cos(\Delta\omega t) = -1$, $A_{\text{mod}} = A[1 – 2] = -A$, giving a secondary intensity peak of $I = A^2$, but the absolute minimum is zero).
• Intensity Ratio: Since $I_{\min} = 0$, the maximum-to-minimum contrast is theoretically infinite:
$$\mathbf{\frac{I_{\max}}{I_{\min}} \to \infty}$$
4. Repetition Frequency of Main Intensity Peaks: The envelope term $\cos(\Delta\omega t)$ repeats with angular frequency $\Delta\omega = 2\pi \Delta f$. The primary maximum ($I = 9A^2$) occurs once per cycle of $\Delta\omega t = 2m\pi$. Hence, the number of primary beats heard per second is:
$$\mathbf{f_b = \frac{\Delta\omega}{2\pi} = \Delta f = |f_2 – f_1| = |f_3 – f_2|}$$
Example 4: Doppler-Coupled Wall Reflection & Beats (JEE Advanced Mastery)
Problem Statement: A sound source producing a pure tone of frequency $f_0 = 660\text{ Hz}$ moves at a constant velocity $u = 5.0\text{ m/s}$ directly toward a rigid, immovable vertical wall. A stationary observer stands at a distance directly behind the moving source (such that the source moves along the line joining the observer and the wall, heading away from the observer and toward the wall). Speed of sound in still air is $v = 330.0\text{ m/s}$.
(a) Determine the frequency $f_{\text{direct}}$ of the sound waves reaching the observer directly from the source.
(b) Determine the frequency $f_{\text{echo}}$ of the reflected waves reaching the observer from the wall.
(c) Find the beat frequency $f_b$ detected by the observer.
Step-by-Step Solution:
1. Part (a) — Direct Wave Frequency ($f_{\text{direct}}$):
The source moves directly away from the stationary observer with velocity $u = 5.0\text{ m/s}$. Applying the Doppler formula for a receding source:
$$f_{\text{direct}} = f_0 \left(\frac{v}{v + u}\right) = 660 \left(\frac{330.0}{330.0 + 5.0}\right) = 660 \left(\frac{330.0}{335.0}\right) = 660 \times \frac{66}{67} = \frac{43560}{67} \approx \mathbf{650.15\text{ Hz}}$$
2. Part (b) — Reflected Wave Frequency ($f_{\text{echo}}$):
The rigid wall acts first as a stationary observer receiving sound from an approaching source:
$$f_{\text{wall}} = f_0 \left(\frac{v}{v – u}\right) = 660 \left(\frac{330.0}{330.0 – 5.0}\right) = 660 \left(\frac{330.0}{325.0}\right) = 660 \times \frac{66}{65} = \frac{43560}{65} = \mathbf{670.15\text{ Hz}}$$
The wall reflects this incident frequency without modification. To the stationary observer, the wall acts as a stationary source emitting frequency $f_{\text{echo}} = f_{\text{wall}} = 670.15\text{ Hz}$.
3. Part (c) — Beat Frequency Heard by Observer:
The observer receives both $f_{\text{echo}}$ and $f_{\text{direct}}$ simultaneously. The resulting beat frequency is:
$$f_b = f_{\text{echo}} – f_{\text{direct}} = f_0 \left(\frac{v}{v – u} – \frac{v}{v + u}\right) = f_0 v \left[\frac{(v + u) – (v – u)}{v^2 – u^2}\right] = \mathbf{\frac{2 u v f_0}{v^2 – u^2}}$$
Substituting the numerical values:
$$f_b = \frac{2 \times 5.0 \times 330.0 \times 660}{(330.0)^2 – (5.0)^2} = \frac{2178000}{108900 – 25} = \frac{2178000}{108875} \approx \mathbf{20.00\text{ Hz}}$$
Analytical Check using $u \ll v$:
$$f_b \approx \frac{2u}{v} f_0 = \frac{2(5.0)}{330.0} \times 660 = \frac{10}{330} \times 660 = 20\text{ Hz}$$
Hence, the beat frequency is exactly $\mathbf{20\text{ Hz}}$.
Practice Worksheet: 10 NTA-Pattern Questions
Section A: Single Correct Questions (Q1 to Q5)
Q1. A tuning fork of unknown frequency produces $5\text{ beats/second}$ when sounded with a fork of frequency $250\text{ Hz}$. When the unknown fork is loaded with a small amount of wax, the beat frequency becomes $2\text{ beats/second}$. The original frequency of the unknown fork was:
(A) $245\text{ Hz}$
(B) $248\text{ Hz}$
(C) $252\text{ Hz}$
(D) $255\text{ Hz}$
Q2. Two vibrating tuning forks produce sound displacement waves given by $s_1 = 4\sin(500\pi t)$ and $s_2 = 2\sin(506\pi t)$ (where $s$ is in millimeters and $t$ is in seconds). The number of beats heard per second and the ratio of maximum to minimum sound intensity are respectively:
(A) $3\text{ beats/s}$, $9 : 1$
(B) $6\text{ beats/s}$, $9 : 1$
(C) $3\text{ beats/s}$, $4 : 1$
(D) $6\text{ beats/s}$, $3 : 1$
Q3. A closed organ pipe of length $L_1$ and an open organ pipe of length $L_2$ both vibrate in their fundamental modes, producing $4\text{ beats/second}$. If the length of the open pipe is halved and that of the closed pipe is doubled, the new beat frequency between their fundamental modes (in air at the same temperature) will be:
(A) $4\text{ beats/s}$
(B) $8\text{ beats/s}$
(C) Dependent on the original lengths, cannot be determined from given data alone
(D) $16\text{ beats/s}$
Q4. In an experiment to determine the speed of sound, two open organ pipes of lengths $50.0\text{ cm}$ and $50.5\text{ cm}$ produce $3\text{ beats/second}$ when sounded together in their fundamental modes. Neglecting end corrections, the speed of sound in air is:
(A) $300\text{ m/s}$
(B) $303\text{ m/s}$
(C) $330\text{ m/s}$
(D) $340\text{ m/s}$
Q5. Two sound waves having frequencies $f_1 = 300\text{ Hz}$ and $f_2 = 304\text{ Hz}$ interfere to produce beats. The time interval between a condition of maximum intensity and the immediately succeeding condition of minimum intensity is:
(A) $0.250\text{ s}$
(B) $0.125\text{ s}$
(C) $0.0625\text{ s}$
(D) $0.500\text{ s}$
Section B: Multi-Correct Questions (Q6 to Q8)
Q6. Two collinear sound waves of frequencies $f_1$ and $f_2$ ($f_1 > f_2$) and equal amplitudes $A$ superpose at a point. Which of the following statements is/are correct?
(A) The frequency of the modulated amplitude envelope is $\frac{f_1 – f_2}{2}$.
(B) The frequency of intensity variation (beat frequency) is $f_1 – f_2$.
(C) The carrier oscillation frequency of medium particles is $\frac{f_1 + f_2}{2}$.
(D) The maximum intensity is four times the intensity of either individual wave.
Q7. Tuning fork $A$ of frequency $f_A$ produces $b$ beats per second when sounded with tuning fork $B$ of frequency $f_B$. Which of the following deductions is/are logically sound?
(A) If loading fork $A$ with wax decreases the beat frequency, then $f_A > f_B$.
(B) If loading fork $A$ with wax increases the beat frequency, then $f_A < f_B$.
(C) If filing fork $A$ decreases the beat frequency, then $f_A < f_B$.
(D) If filing fork $A$ increases the beat frequency, then $f_A > f_B$.
Q8. Three harmonic sound waves given by $s_1 = A\sin(2\pi \cdot 400 t)$, $s_2 = A\sin(2\pi \cdot 402 t)$, and $s_3 = A\sin(2\pi \cdot 404 t)$ superpose at a detector. Which of the following statements is/are correct?
(A) The carrier frequency of particle vibration is $402\text{ Hz}$.
(B) The maximum possible displacement amplitude is $3A$.
(C) The number of primary intensity maxima per second is $2$.
(D) The minimum intensity during each beat cycle is strictly zero.
Section C: Numerical Value Questions (Q9 to Q10)
Q9. A set of 25 tuning forks is arranged in ascending order of their frequencies. Each fork produces $4\text{ beats/second}$ when sounded with the fork immediately preceding it. If the frequency of the last fork is exactly twice that of the first fork (an octave higher), find the frequency of the first fork (in $\text{Hz}$).
Q10. A wire of length $L = 1.0\text{ m}$ under tension $T$ vibrates in its fundamental mode and produces $5\text{ beats/second}$ with a standard tuning fork. When the length of the wire is reduced to $0.98\text{ m}$ without changing the tension, it again produces $5\text{ beats/second}$ with the same tuning fork. Find the frequency of the tuning fork (in $\text{Hz}$, rounded to the nearest integer).
Comprehensive Solutions & Answer Key
Answer Key:
Q1: (D) Q2: (A) Q3: (C) Q4: (B) Q5: (B)
Q6: (A, B, C, D) Q7: (A, B, C, D) Q8: (A, B, C, D)
Q9: 96 Q10: 495
Solution to Q1:
Given $f_B = 250\text{ Hz}$ and initial beat frequency $b = 5\text{ Hz}$.
The possible initial frequencies for fork $A$ are:
$$f_A = 250 \pm 5 \implies f_A = 255\text{ Hz} \quad \text{OR} \quad f_A = 245\text{ Hz}$$
Loading fork $A$ with wax decreases its frequency ($f_A’ < f_A$).
• If $f_A = 255\text{ Hz}$, decreasing $f_A$ moves it closer to $250\text{ Hz}$, reducing the beat frequency to $2\text{ Hz}$ ($f_A’ = 252\text{ Hz}$). This matches the observation.
• If $f_A = 245\text{ Hz}$, decreasing $f_A$ moves it further below $250\text{ Hz}$, which would increase the beat frequency ($|f_A’ – 250| > 5$).
Hence, the original frequency is uniquely $\mathbf{255\text{ Hz}}$. Choice (D) is correct.
Solution to Q2:
Comparing with $s = A\sin(2\pi f t)$:
Wave 1: $2\pi f_1 = 500\pi \implies f_1 = 250\text{ Hz}$, with amplitude $A_1 = 4\text{ mm}$.
Wave 2: $2\pi f_2 = 506\pi \implies f_2 = 253\text{ Hz}$, with amplitude $A_2 = 2\text{ mm}$.
1. Beat frequency:
$$f_b = |f_2 – f_1| = 253 – 250 = \mathbf{3\text{ beats/second}}$$
2. Ratio of maximum to minimum intensity:
$$A_{\max} = A_1 + A_2 = 4 + 2 = 6\text{ mm}$$
$$A_{\min} = A_1 – A_2 = 4 – 2 = 2\text{ mm}$$
$$\frac{I_{\max}}{I_{\min}} = \left(\frac{A_{\max}}{A_{\min}}\right)^2 = \left(\frac{6}{2}\right)^2 = 3^2 = \mathbf{9 : 1}$$
Hence, choice (A) is correct.
Solution to Q3:
Original frequencies:
$$f_{c} = \frac{v}{4L_1}, \quad f_{o} = \frac{v}{2L_2}, \quad \text{with } |f_c – f_o| = 4\text{ Hz}$$
When $L_o’ = L_2/2$, the new open pipe frequency is:
$$f_o’ = \frac{v}{2(L_2/2)} = \frac{v}{L_2} = 2f_o$$
When $L_c’ = 2L_1$, the new closed pipe frequency is:
$$f_c’ = \frac{v}{4(2L_1)} = \frac{v}{8L_1} = \frac{f_c}{2}$$
The new beat frequency is $f_b’ = |f_o’ – f_c’| = |2f_o – f_c/2|$.
Since $|f_o – f_c| = 4$ leaves the individual absolute values of $f_o$ and $f_c$ undetermined, the combination $|2f_o – f_c/2|$ cannot be uniquely evaluated from the beat frequency alone. Choice (C) is correct.
Solution to Q4:
For an open organ pipe, $f = \frac{v}{2L}$.
Given $L_1 = 50.0\text{ cm} = 0.500\text{ m}$ and $L_2 = 50.5\text{ cm} = 0.505\text{ m}$.
The beat frequency is:
$$f_b = f_1 – f_2 = \frac{v}{2}\left(\frac{1}{L_1} – \frac{1}{L_2}\right) = \frac{v}{2}\left(\frac{L_2 – L_1}{L_1 L_2}\right)$$
Substitute given values ($f_b = 3\text{ Hz}$, $L_2 – L_1 = 0.005\text{ m}$):
$$3 = \frac{v}{2} \left[\frac{0.005}{0.500 \times 0.505}\right] = \frac{v}{2} \left[\frac{0.005}{0.2525}\right] = \frac{v}{2} \left(\frac{1}{50.5}\right) = \frac{v}{101}$$
$$v = 3 \times 101 = \mathbf{303\text{ m/s}}$$
Hence, choice (B) is correct.
Solution to Q5:
The beat frequency is:
$$f_b = |f_2 – f_1| = 304 – 300 = 4\text{ Hz}$$
The time period of one complete beat (maximum to maximum) is:
$$T_b = \frac{1}{f_b} = \frac{1}{4} = 0.250\text{ s}$$
The time interval from maximum intensity to the immediately succeeding minimum intensity is half of the beat period:
$$\mathbf{\Delta t = \frac{T_b}{2} = \frac{0.250\text{ s}}{2} = 0.125\text{ s}}$$
Hence, choice (B) is correct.
Solution to Q6:
Let us evaluate each statement:
(A) The amplitude envelope is $A_{\text{mod}}(t) = 2A\cos[\pi(f_1 – f_2)t]$. The angular frequency is $\pi(f_1 – f_2) = 2\pi f_{\text{envelope}} \implies f_{\text{envelope}} = \frac{f_1 – f_2}{2}$. (A) is correct.
(B) Since intensity is proportional to amplitude squared, $I \propto \cos^2[\pi(f_1 – f_2)t] = \frac{1 + \cos[2\pi(f_1 – f_2)t]}{2}$. The frequency of intensity variation is $f_1 – f_2$. (B) is correct.
(C) The carrier oscillation term is $\sin\left(2\pi \frac{f_1 + f_2}{2} t\right)$, so the carrier frequency is $\frac{f_1 + f_2}{2}$. (C) is correct.
(D) $I_{\max} \propto (A + A)^2 = 4A^2 = 4I_0$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
(A) Loading decreases $f_A$. If $f_A > f_B$, decreasing $f_A$ brings it closer to $f_B$, reducing $b$. (A) is correct.
(B) If $f_A < f_B$, decreasing $f_A$ widens the gap, increasing $b$. (B) is correct.
(C) Filing increases $f_A$. If $f_A < f_B$, increasing $f_A$ narrows the gap, decreasing $b$. (C) is correct.
(D) If $f_A > f_B$, increasing $f_A$ widens the gap, increasing $b$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
Here $f_1 = 400\text{ Hz}$, $f_2 = 402\text{ Hz}$, $f_3 = 404\text{ Hz}$. They are in arithmetic progression with $f_0 = 402\text{ Hz}$ and $\Delta f = 2\text{ Hz}$.
(A) The resultant displacement is $s(t) = A[1 + 2\cos(2\pi \cdot 2 t)]\sin(2\pi \cdot 402 t)$. The carrier frequency is $402\text{ Hz}$. (A) is correct.
(B) When $\cos(4\pi t) = 1$, $A_{\max} = A(1 + 2) = 3A$. (B) is correct.
(C) The envelope modulates at $\Delta f = 2\text{ Hz}$, producing 2 primary intensity peaks per second. (C) is correct.
(D) When $\cos(4\pi t) = -1/2$, $A_{\text{mod}} = A(1 – 1) = 0$, so intensity drops strictly to zero. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q9:
Let the frequencies of the 25 forks be $f_1, f_2, \dots, f_{25}$ in ascending order.
Each fork produces $4\text{ beats/s}$ with the preceding one:
$$f_2 = f_1 + 4$$
$$f_3 = f_2 + 4 = f_1 + 2(4)$$
$$f_{25} = f_1 + (25 – 1) \times 4 = f_1 + 24 \times 4 = f_1 + 96$$
Given that the last fork is an octave higher than the first ($f_{25} = 2f_1$):
$$2f_1 = f_1 + 96 \implies \mathbf{f_1 = 96\text{ Hz}}$$
Answer: 96
Solution to Q10:
The fundamental frequency of a stretched wire of length $L$ is $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \propto \frac{1}{L}$.
Let $f_0$ be the frequency of the tuning fork.
When length is $L_1 = 1.0\text{ m}$, the frequency is $f_1 \propto 1/1.0$.
When length is reduced to $L_2 = 0.98\text{ m}$, the wire frequency increases: $f_2 > f_1$.
Since both states produce $5\text{ beats/second}$ with $f_0$:
$$f_1 = f_0 – 5 \quad \text{and} \quad f_2 = f_0 + 5$$
Taking the ratio of the two frequencies:
$$\frac{f_2}{f_1} = \frac{L_1}{L_2} = \frac{1.0}{0.98} = \frac{100}{98} = \frac{50}{49}$$
Substitute expressions in terms of $f_0$:
$$\frac{f_0 + 5}{f_0 – 5} = \frac{50}{49}$$
Cross-multiplying:
$$49(f_0 + 5) = 50(f_0 – 5)$$
$$49 f_0 + 245 = 50 f_0 – 250$$
$$50 f_0 – 49 f_0 = 245 + 250$$
$$\mathbf{f_0 = 495\text{ Hz}}$$
Answer: 495