Q.When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?

Showing the 12 most recent of 37 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A source of sound of frequency 660 Hz and an observer are moving towards each other with speeds of 31 kmph and 23 kmph respectively. If the wind blows with a speed of 5 kmph from observer towards the source, then the frequency of the sound heard by the observer is (Speed of sound in air $= 340~\mathrm{ms}^{-1}$) (A) 630 Hz (B) 660 Hz (C) 690 Hz (D) 720 Hz

   ›Reveal solutionSolution

   The presence of wind means we must calculate the speeds of the source and observer relative to the air (the medium) before applying the Doppler effect formula. The observed frequency is $\boxed{690~\mathrm{Hz}}$.

   Concept and Intuition

   The Doppler effect describes the change in frequency of a wave for an observer moving relative to its source. For sound waves, this effect depends on the relative velocities of the source, the observer, and the medium (air) through which the sound travels.

   The crucial point when wind is present is that the speed of sound ($v$) in the Doppler formula refers to the speed of sound relative to the medium (i.e., in still air). Similarly, the velocities of the source ($v_S$) and the observer ($v_O$) must also be measured relative to the medium. The wind effectively changes the velocities of the source and observer with respect to the air, even if their velocities relative to the ground remain constant.

   We will first convert all given speeds to a consistent unit (m/s). Then, we will determine the velocities of the source and observer relative to the air by accounting for the wind's velocity. Finally, we will use the standard Doppler effect formula for sound, ensuring the correct signs for relative motion.

   The general Doppler effect formula for sound, where $v$ is the speed of sound in the medium, $v_O$ is the speed of the observer relative to the medium, and $v_S$ is the speed of the source relative to the medium, is:

   $f' = f_0 \left( \frac{v \pm v_O}{v \mp v_S} \right)$

   Here, we use $+v_O$ if the observer moves towards the source and $-v_O$ if the observer moves away. We use $-v_S$ if the source moves towards the observer and $+v_S$ if the source moves away.

   Step-by-step Derivations

   1. Convert all speeds to meters per second (m/s):

      The speed of sound in air is given as $v = 340~\mathrm{m/s}$.

      The speeds of the source, observer, and wind are given in kmph. We convert them to m/s using the conversion factor $1~\mathrm{kmph} = \frac{5}{18}~\mathrm{m/s}$.

      • Speed of source relative to ground, $v_{S,g} = 31~\mathrm{kmph} = 31 \times \frac{5}{18}~\mathrm{m/s} = \frac{155}{18}~\mathrm{m/s}$.
      • Speed of observer relative to ground, $v_{O,g} = 23~\mathrm{kmph} = 23 \times \frac{5}{18}~\mathrm{m/s} = \frac{115}{18}~\mathrm{m/s}$.
      • Speed of wind relative to ground, $v_{W,g} = 5~\mathrm{kmph} = 5 \times \frac{5}{18}~\mathrm{m/s} = \frac{25}{18}~\mathrm{m/s}$.

   2. Establish a coordinate system and determine velocities relative to the ground:

      Let's define the direction from the source to the observer as the positive direction.

      • The source is moving towards the observer, so its velocity relative to the ground is in the negative direction: $V_{S,g} = -\frac{155}{18}~\mathrm{m/s}$.
      • The observer is moving towards the source, so its velocity relative to the ground is in the positive direction: $V_{O,g} = +\frac{115}{18}~\mathrm{m/s}$.
      • The wind blows from the observer towards the source. This means the wind is blowing in the negative direction: $V_{W,g} = -\frac{25}{18}~\mathrm{m/s}$.

   3. Calculate the velocities of the source and observer relative to the medium (air):

      The velocity of an object relative to the medium is given by $V_{\text{object, air}} = V_{\text{object, ground}} - V_{\text{wind, ground}}$.

      • Velocity of source relative to air ($V_{S,air}$):

        $V_{S,air} = V_{S,g} - V_{W,g} = \left(-\frac{155}{18}\right) - \left(-\frac{25}{18}\right) = -\frac{155}{18} + \frac{25}{18} = -\frac{130}{18}~\mathrm{m/s} = -\frac{65}{9}~\mathrm{m/s}$.

        This means the source is moving at a speed of $\frac{65}{9}~\mathrm{m/s}$ towards the observer relative to the air. So, $v_S = \frac{65}{9}~\mathrm{m/s}$.

      • Velocity of observer relative to air ($V_{O,air}$):

        $V_{O,air} = V_{O,g} - V_{W,g} = \left(+\frac{115}{18}\right) - \left(-\frac{25}{18}\right) = +\frac{115}{18} + \frac{25}{18} = +\frac{140}{18}~\mathrm{m/s} = +\frac{70}{9}~\mathrm{m/s}$. …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Two stationary sources P and Q produce sounds of equal frequency of 170 Hz. The velocity with which an observer has to move from source P towards source Q such that 8 beats are to be heard per second by the observer is (Speed of sound in air $= 340~\mathrm{ms}^{-1}$) (A) $20~\mathrm{ms}^{-1}$ (B) $16~\mathrm{ms}^{-1}$ (C) $4~\mathrm{ms}^{-1}$ (D) $8~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Moving away from P and toward Q, the beat count is $f\dfrac{2v_o}{c}=8$, giving $v_o=8~\mathrm{m\,s^{-1}}$.

   An observer moving with speed $v_o$ from P toward Q recedes from source P and approaches source Q. The frequencies heard:

   $f_P=f\,\frac{c-v_o}{c},\qquad f_Q=f\,\frac{c+v_o}{c}.$

   Beat frequency:

   $f_Q-f_P=f\,\frac{2v_o}{c}=8.$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Two observers A and B are moving towards a stationary source of sound with speeds $0.8V_0$ and $0.6V_0$ respectively, where $V_0$ is the speed of sound in air. The ratio of the frequencies of the sound heard by the observers A and B is (A) $4 : 3$ (B) $9 : 8$ (C) $1 : 1$ (D) $2 : 3$

   ›Reveal solutionSolution

   The Doppler effect for a moving observer and stationary source gives $f' = f_0 \left(1 + \frac{v_o}{V_0}\right)$. Substituting the given speeds yields a ratio of $9:8$ for observer A to observer B.

   The core idea here is the Doppler effect — the change in observed frequency when there is relative motion between source and observer. Since the source is stationary and both observers are moving toward it, the sound waves get "compressed" from the observer's perspective, raising the pitch. The faster the observer moves, the greater the frequency shift.

   For a stationary source and a moving observer, the formula is straightforward: the observed frequency $f'$ equals the source frequency $f_0$ multiplied by $(1 + v_o/V_0)$ when the observer moves toward the source. Here $v_o$ is the observer's speed and $V_0$ is the speed of sound. No sign confusion — just add the ratio.

   Let’s work it out step by step.

   1. Write the Doppler formula for a moving observer, stationary source. When the observer moves toward the source:

   $f' = f_0 \left( \frac{V_0 + v_o}{V_0} \right) = f_0 \left(1 + \frac{v_o}{V_0}\right)$

   This is because the observer encounters wavefronts more frequently than a stationary observer would.

   2. Apply to observer A. Speed of A: $v_A = 0.8V_0$.

   $f_A = f_0 \left(1 + \frac{0.8V_0}{V_0}\right) = f_0 (1 + 0.8) = 1.8 f_0$

   3. Apply to observer B. Speed of B: $v_B = 0.6V_0$. $f_B = f_0 \left(1 + \frac{0.6V_0}{V_0}\right) = f_0 (1 + 0.6) = 1.6 f_0$ …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The approximate bandwidth of frequency required to transmit music is (A) 20 kHz (B) 2800 Hz (C) 4.2 MHz (D) 100 MHz

   ›Reveal solutionSolution

   The bandwidth needed to transmit music is determined by the highest audible frequency, which is about 20 kHz. Since bandwidth equals the highest frequency in a baseband signal, the correct choice is (A) 20 kHz.

   The key concept here is bandwidth in the context of audio signals. For a signal like music, the bandwidth is simply the range of frequencies it contains. Human hearing typically spans from about 20 Hz to 20 kHz. To transmit music without significant loss of quality, the transmission system must accommodate this entire range. The lowest frequency (20 Hz) is negligible compared to the highest, so the bandwidth is essentially the highest frequency present — roughly 20 kHz.

   Let’s walk through the reasoning step by step:

   1. Understand what bandwidth means for audio.

      Bandwidth is defined as the difference between the highest and lowest frequencies in a signal. For music, the lowest audible frequency is around 20 Hz, and the highest is around 20,000 Hz (20 kHz). So the raw bandwidth is $20\,000 - 20 = 19\,980$ Hz, which rounds to 20 kHz.

   2. Why not 2800 Hz?

      2800 Hz (2.8 kHz) is roughly the bandwidth of a telephone line, which is sufficient for speech but far too narrow for music. Music contains harmonics and overtones that extend well beyond 2.8 kHz; cutting off at that frequency would make music sound muffled and dull.

   3. Why not 4.2 MHz or 100 MHz? …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A source producing sound of frequency 990 Hz and an observer are initially at rest. If both the source and observer start moving simultaneously towards each other with accelerations $2\,\mathrm{ms}^{-2}$ and $4\,\mathrm{ms}^{-2}$ respectively, then the frequency of sound heard by the observer at a time of $t = 5\,\mathrm{s}$ is (Speed of sound in air $= 340\,\mathrm{ms}^{-1}$) (A) 960 Hz (B) 1080 Hz (C) 1050 Hz (D) 900 Hz

   ›Reveal solutionSolution

   Both source and observer accelerate toward each other; at $t=5\,\text{s}$ their speeds are $v_s=10\,\text{m/s}$ and $v_o=20\,\text{m/s}$, giving $f'=1080\,\text{Hz}$.

   At $t=5\,\text{s}$ the speeds acquired from rest are:

   • Source: $v_s = a_s t = 2\times 5 = 10\,\text{m/s}$ (moving toward the observer)
   • Observer: $v_o = a_o t = 4\times 5 = 20\,\text{m/s}$ (moving toward the source) …

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6. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The frequency of a tuning fork P is 1.5% more than the frequency of tuning fork Q and the frequency of another tuning fork R is 2.5% less than the frequency of tuning fork Q. If 8 beats are produced per second when P and R are vibrated together, then the frequency of tuning fork R is (A) 203 Hz (B) 195 Hz (C) 200 Hz (D) 187 Hz

   ›Reveal solutionSolution

   The problem involves percentage differences relative to a common reference (Q). Let Q’s frequency be $f$, express P and R in terms of $f$, use the beat frequency condition $|f_P - f_R| = 8$ Hz, solve for $f$, then find $f_R$. The answer is 195 Hz.

   The core idea is that beats per second equal the absolute difference in frequencies. When two tuning forks are sounded together, the number of beats you hear each second is simply $|f_1 - f_2|$. Here, P and R produce 8 beats per second, so $|f_P - f_R| = 8$.

   Both P and R are described relative to Q. That makes Q the natural reference. Let the frequency of tuning fork Q be $f$ Hz. Then:

   1. P is 1.5% more than Q. That means $f_P = f + 1.5\% \text{ of } f = f + 0.015f = 1.015f$.

   2. R is 2.5% less than Q. So $f_R = f - 2.5\% \text{ of } f = f - 0.025f = 0.975f$.

   Now, P and R together give 8 beats per second. Since P has a higher frequency than R (because 1.015 > 0.975), the difference is $f_P - f_R = 8$ Hz. No absolute value needed — we know which is larger.

   So:

   $1.015f - 0.975f = 8$

   $0.04f = 8$

   $f = \frac{8}{0.04} = 200 \text{ Hz}$

   That is the frequency of Q. The question asks for the frequency of R:

   $f_R = 0.975 \times 200 = 195 \text{ Hz}$ …

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7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A tuning fork P of frequency 384 Hz produces 6 beats per second with a tuning fork Q. When a little wax is attached to Q and again P and Q are sounded together, if the number of beats produced per second do not change, then the initial frequency of Q is (A) 390 Hz (B) 378 Hz (C) 381 Hz (D) 387 Hz

   ›Reveal solutionSolution

   The beat frequency equals the absolute difference of the two frequencies. Adding wax lowers Q’s frequency. Since the beat count stays the same after waxing, Q’s initial frequency must have been higher than P’s, so the answer is 390 Hz.

   The concept here is beats — the periodic variation in loudness when two sound waves of slightly different frequencies interfere. The beat frequency is simply $|f_P - f_Q|$. The trick in this problem is to think about what happens when you add wax to a tuning fork: wax increases its mass, which lowers its natural frequency. So after waxing, Q’s frequency decreases.

   Now, the beats per second remain 6 both before and after waxing. That means the absolute difference between P and Q’s frequencies stays 6 Hz, even though Q’s frequency has dropped. Let’s work through the possibilities.

   1. Before waxing: $|384 - f_Q| = 6$. So $f_Q$ could be either $384 + 6 = 390$ Hz or $384 - 6 = 378$ Hz. Both are possible initially.

   2. After waxing: Q’s frequency becomes $f_Q'$, which is less than $f_Q$ (because wax lowers it). The beat frequency is still 6, so $|384 - f_Q'| = 6$.

   3. Test the two cases:

      • If $f_Q = 378$ Hz initially, then after waxing $f_Q' < 378$ Hz. The difference $|384 - f_Q'|$ would then be greater than 6 (since 384 is fixed, and Q moves further away). So beats would increase — not stay the same. This case fails. …

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Two tuning forks of frequencies 320 Hz and 323 Hz are vibrated together. The time interval between a maximum sound and its adjacent minimum sound heard by an observer is (A) $\dfrac{1}{6}$ s (B) $\dfrac{1}{3}$ s (C) $\dfrac{1}{12}$ s (D) $\dfrac{1}{9}$ s

   ›Reveal solutionSolution

   When two close frequencies are sounded together, they produce beats. The time between a maximum and the adjacent minimum is half the beat period. Here the beat frequency is 3 Hz, so that time is 1/6 s.

   Concept & Intuition

   When two tuning forks with slightly different frequencies vibrate together, the sound you hear fluctuates in loudness. This is called beats. The loudness rises to a maximum when the two waves are in phase (constructive interference) and falls to a minimum when they are out of phase (destructive interference). The number of maxima per second is the beat frequency, which equals the absolute difference of the two frequencies. The time from one maximum to the next maximum is the beat period $T_{\text{beat}} = 1/f_{\text{beat}}$. But the question asks for the time from a maximum to the adjacent minimum — that’s exactly half a beat period, because the pattern goes max → min → max in one full cycle.

   Step-by-step solution

   1. Find the beat frequency The two frequencies are $f_1 = 320\ \text{Hz}$ and $f_2 = 323\ \text{Hz}$. Beat frequency:

   $f_{\text{beat}} = |f_2 - f_1| = 323 - 320 = 3\ \text{Hz}.$

   2. Find the beat period The beat period is the time between successive maxima: $$$$

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Two tuning forks of frequencies 320 Hz and 323 Hz are vibrated together. The time interval between a maximum sound and its adjacent minimum sound heard by an observer is (A) $\dfrac{1}{6}$ s (B) $\dfrac{1}{3}$ s (C) $\dfrac{1}{9}$ s (D) $\dfrac{1}{12}$ s

   ›Reveal solutionSolution

   When two close frequencies are sounded together, they produce beats. The time between a maximum and the adjacent minimum is half the beat period. Here the beat frequency is 3 Hz, so that interval is 1/6 s.

   Concept & Intuition

   When two tuning forks with slightly different frequencies vibrate together, the sound you hear fluctuates in loudness. These fluctuations are called beats. The beat frequency equals the absolute difference of the two source frequencies:

   $f_{\text{beat}} = |f_1 - f_2|.$

   The beat period is the time between successive maxima (or successive minima) of loudness:

   $T_{\text{beat}} = \frac{1}{f_{\text{beat}}}.$

   But the question asks for the time from a maximum to the adjacent minimum. Since the loudness pattern is sinusoidal, a maximum and the next minimum are separated by exactly half a beat period.

   ---

   Step-by-step solution

   1. Find the beat frequency

   $f_1 = 320\ \text{Hz},\quad f_2 = 323\ \text{Hz}$

   $f_{\text{beat}} = |323 - 320| = 3\ \text{Hz}.$

   2. Find the beat period $T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{1}{3}\ \text{s}.$ …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.If the frequencies of the carrier wave and message signal are 1 MHz and 28 kHz respectively, then the frequencies of the side bands are (A) 1014 kHz, 986 kHz (B) 1028 kHz, 972 kHz (C) 29 kHz, 27 kHz (D) 514 kHz, 486 kHz

    ›Reveal solutionSolution

    In amplitude modulation, the sideband frequencies are the sum and difference of the carrier and message frequencies. With a 1 MHz carrier and a 28 kHz message, the sidebands are 1028 kHz and 972 kHz, which matches option (B).

    The key idea is that amplitude modulation (AM) does not simply shift the message to the carrier frequency — it creates two copies of the message spectrum, one above and one below the carrier. These are called the upper sideband (USB) and lower sideband (LSB). Their frequencies are given by:

    $f_{\text{USB}} = f_c + f_m \quad \text{and} \quad f_{\text{LSB}} = f_c - f_m$

    where $f_c$ is the carrier frequency and $f_m$ is the message (modulating) frequency.

    Let’s work through it step by step.

    1. Identify the given values

       Carrier frequency: $f_c = 1\ \text{MHz} = 1000\ \text{kHz}$

       Message frequency: $f_m = 28\ \text{kHz}$

    2. Compute the upper sideband frequency

    $f_{\text{USB}} = f_c + f_m = 1000\ \text{kHz} + 28\ \text{kHz} = 1028\ \text{kHz}$

    3. Compute the lower sideband frequency

    $f_{\text{LSB}} = f_c - f_m = 1000\ \text{kHz} - 28\ \text{kHz} = 972\ \text{kHz}$

    4. Match with the options …

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.If the frequencies of the carrier wave and message signal are 1 MHz and 28 kHz respectively, then the frequencies of the side bands are (A) 1028 kHz, 972 kHz (B) 29 kHz, 27 kHz (C) 1014 kHz, 986 kHz (D) 514 kHz, 486 kHz

    ›Reveal solutionSolution

    In amplitude modulation, the sideband frequencies are the sum and difference of the carrier and message frequencies. With a 1 MHz carrier and a 28 kHz message, the sidebands are 1028 kHz and 972 kHz, so option (A) is correct.

    Concept & Intuition

    Amplitude modulation (AM) works by varying the amplitude of a high‑frequency carrier wave in proportion to a lower‑frequency message signal. Mathematically, multiplying the carrier by the message produces new frequencies: the carrier itself plus two “sidebands” — one above the carrier (upper sideband) and one below (lower sideband). The upper sideband is at $f_c + f_m$, the lower at $f_c - f_m$. This is a direct consequence of the trigonometric identity for the product of two sinusoids.

    Step‑by‑step reasoning

    1. Identify the given frequencies

       Carrier frequency: $f_c = 1\ \text{MHz} = 1000\ \text{kHz}$

       Message (modulating) frequency: $f_m = 28\ \text{kHz}$

    2. Recall the sideband formula

       For standard AM (double‑sideband, full carrier), the sideband frequencies are:

    $f_{\text{USB}} = f_c + f_m, \qquad f_{\text{LSB}} = f_c - f_m$

    3. Compute the upper sideband

    $f_{\text{USB}} = 1000\ \text{kHz} + 28\ \text{kHz} = 1028\ \text{kHz}$

    4. Compute the lower sideband $$$$

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12. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.A wire under a tension of 144 N vibrating in its fundamental mode gives 5 beats per second with a tuning fork. When the tension applied to the wire is increased to 169 N, the number of beats heard per second remains the same. The frequency of the tuning fork is (A) $125\ \text{Hz}$ (B) $60\ \text{Hz}$ (C) $65\ \text{Hz}$ (D) $55\ \text{Hz}$

    ›Reveal solutionSolution

    The frequency of a vibrating wire is proportional to the square root of its tension. Since the beat frequency with a tuning fork remains constant when the wire's tension increases, the tuning fork's frequency must lie between the two wire frequencies. This leads to the tuning fork's frequency being $\boxed{125\ \text{Hz}}$.

    The problem describes a vibrating wire producing beats with a tuning fork under two different tensions. To solve this, we need to understand how the frequency of a vibrating wire depends on tension and what beats signify.

    Concept and Intuition

    1. Frequency of a Vibrating Wire: For a wire vibrating in its fundamental mode, its frequency ($f$) is determined by its length ($L$), tension ($T$), and linear mass density ($\mu$). The formula is:

       $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

       For a specific wire, $L$ and $\mu$ are constant. This means the frequency of the wire is directly proportional to the square root of the tension: $f \propto \sqrt{T}$.

       Therefore, if the tension increases, the frequency of the wire will also increase.

    2. Beats: When two sound waves of slightly different frequencies ($f_1$ and $f_2$) interfere, they produce beats. The beat frequency ($f_{beat}$) is the absolute difference between their frequencies:

       $f_{beat} = |f_1 - f_2|$

       In this problem, the two frequencies are that of the vibrating wire ($f_w$) and the tuning fork ($f_t$). So, $f_{beat} = |f_w - f_t|$.

    3. Combining the Concepts: We are given that the beat frequency remains the same (5 Hz) even when the tension in the wire is increased. Since increasing the tension increases the wire's frequency ($f_w$), this implies a specific relationship between $f_w$ and $f_t$.

       Let $f_{w1}$ be the wire's frequency at tension $T_1$ and $f_{w2}$ be the wire's frequency at tension $T_2$.

       Since $T_2 > T_1$, we know $f_{w2} > f_{w1}$.

       The beat frequency is 5 Hz in both cases:

       $|f_{w1} - f_t| = 5$

       $|f_{w2} - f_t| = 5$

       Consider the possibilities for the initial state ($f_{w1}$ relative to $f_t$):

       • Case A: $f_{w1} > f_t$. Then $f_{w1} - f_t = 5$. As tension increases, $f_{w2}$ increases, so $f_{w2} > f_{w1}$. If $f_{w2}$ is still greater than $f_t$, then $f_{w2} - f_t > f_{w1} - f_t = 5$. The beat frequency would increase, which contradicts the problem statement.
       • Case B: $f_{w1} < f_t$. Then $f_t - f_{w1} = 5$. As tension increases, $f_{w2}$ increases, so $f_{w2} > f_{w1}$. If $f_{w2}$ is still less than $f_t$, then $f_t - f_{w2} < f_t - f_{w1} = 5$. The beat frequency would decrease, which contradicts the problem statement.
       • Case C: $f_{w1} < f_t$ and $f_{w2} > f_t$. This is the only scenario where the beat frequency can remain the same. The wire's frequency "crosses over" the tuning fork's frequency.
         • Initially, $f_t - f_{w1} = 5$.
         • Finally, $f_{w2} - f_t = 5$.

       This means the tuning fork's frequency ($f_t$) must lie between the two frequencies of the wire ($f_{w1}$ and $f_{w2}$).

    Step-by-Step Solution

    1. Identify Given Information:

       • Initial tension, $T_1 = 144\ \text{N}$
       • Final tension, $T_2 = 169\ \text{N}$
       • Beat frequency in both cases, $f_{beat} = 5\ \text{Hz}$

    2. Relate Wire Frequencies to Tuning Fork Frequency:

       Let $f_t$ be the frequency of the tuning fork. …

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