Q.Sound waves of wavelength $\lambda$ travelling in a medium with a speed of $v$ m/s enter into another medium where its speed is $2v$ m/s. Wavelength of sound waves in the second medium is

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.In a closed organ pipe, the number of nodes formed in fifth and ninth harmonics are respectively (A) 5, 9 (B) 3, 5 (C) 5, 7 (D) 2, 4

   ›Reveal solutionSolution

   Only odd harmonics exist in a closed organ pipe; counting nodes along the standing wave pattern for the 5th and 9th harmonics gives 3 and 5 respectively. Answer: 3, 5.

   Concept and Intuition

   A closed organ pipe has a node at the closed end (air cannot move there) and an antinode at the open end (air moves freely). Only odd harmonics ($n=1,3,5,7,9,\ldots$) can form standing waves satisfying these boundary conditions, with pipe length $L = n\lambda/4$. Visualizing the standing wave pattern for each harmonic, nodes and antinodes alternate starting from a node at the closed end.

   Step-by-Step Solution

   1. For the $n$th harmonic (n odd) in a closed pipe, the wave pattern has nodes at positions $x=0, \lambda/2, \lambda, \ldots$ and antinodes at $x=\lambda/4, 3\lambda/4, \ldots$ up to length $L=n\lambda/4$.
   2. Counting: nodes occur at every half-wavelength interval starting from the closed end; the total count of nodes for harmonic $n$ is $\dfrac{n+1}{2}$.
   3. For the 5th harmonic ($n=5$): nodes $=\dfrac{5+1}{2}=3$.
   4. For the 9th harmonic ($n=9$): nodes $=\dfrac{9+1}{2}=5$.

   Common Mistakes

   • Assuming harmonic number equals node count directly (that's true for open pipes' overtone counting but not the node-counting here).
   • Forgetting that closed pipes only support odd harmonics, and mislabeling which "harmonic" is meant.

   ✓Final answer

   The correct option is (B) — 3, 5.

   ANSWER: B

2. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.A closed pipe and an open pipe of same length produce 4 beats per second when they are set into vibrations simultaneously. If the lengths of both the pipes were halved, then the number of beats produced per second is (Assume same mode of vibrations in both cases) (A) 8 (B) 4 (C) 1 (D) 2

   ›Reveal solutionSolution

   The beat frequency between a closed pipe and an open pipe of the same length equals the closed pipe's own fundamental frequency; halving the length doubles every frequency, so the beats double from 4 to 8 per second.

   Concept and Intuition

   For a pipe of length $L$ carrying sound of speed $v$:

   • Closed pipe (one end closed) fundamental: $f_c = \dfrac{v}{4L}$
   • Open pipe (both ends open) fundamental: $f_o = \dfrac{v}{2L} = 2f_c$

   So for pipes of equal length, the open pipe's fundamental is always exactly twice the closed pipe's fundamental. All frequencies scale as $1/L$, so halving $L$ doubles every fundamental frequency (and hence any beat frequency built from them).

   Step-by-Step Solution

   1. With length $L$: $f_o - f_c = 2f_c - f_c = f_c = \dfrac{v}{4L}$. This equals the given beat frequency, so $\dfrac{v}{4L} = 4\text{ Hz}$.
   2. With length halved to $L/2$: new closed-pipe fundamental $f_c' = \dfrac{v}{4(L/2)} = \dfrac{v}{2L} = 2f_c = 8\text{ Hz}$. New open-pipe fundamental $f_o' = \dfrac{v}{2(L/2)} = \dfrac{v}{L} = 4f_c = 16\text{ Hz}$.
   3. New beat frequency $= f_o' - f_c' = 16 - 8 = 8\text{ Hz}$.

   (Equivalently: since the beat frequency itself is $f_c$, and $f_c$ doubles when $L$ is halved, the beat frequency simply doubles from 4 to 8 Hz.)

   Common Mistakes

   • Assuming beats simply "double the number of beats" without checking that both pipe frequencies actually scale the same way with $L$.
   • Forgetting that the beat frequency between these two specific pipes always equals the closed pipe's fundamental (a useful shortcut once derived).

   ✓Final answer

   The correct option is (A) — 8.

   ANSWER: A

3. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.If the third harmonic of a closed pipe is in resonance with fourth harmonic of an open pipe, then the ratio of the lengths of the closed and open pipes is (A) 8 : 3 (B) 3 : 8 (C) 3 : 4 (D) 4 : 3

   ›Reveal solutionSolution

   Equating the third-harmonic frequency of a closed pipe with the fourth-harmonic frequency of an open pipe gives a length ratio $L_{closed}:L_{open} = 3:8$.

   Concept and Intuition

   A closed (one end closed) pipe supports only odd harmonics, with the $n$-th harmonic ($n=1,3,5,\dots$) at frequency $f_n = \dfrac{nv}{4L}$. An open (both ends open) pipe supports all harmonics, with the $n$-th harmonic at $f_n = \dfrac{nv}{2L}$. Setting the given harmonic frequencies equal (resonance condition) lets us directly compare the two pipe lengths.

   Step-by-Step Solution

   1. Closed pipe, third harmonic ($n=3$, valid since closed pipes only have odd harmonics):

   $f_c = \frac{3v}{4L_c}$

   2. Open pipe, fourth harmonic ($n=4$):

   $f_o = \frac{4v}{2L_o} = \frac{2v}{L_o}$

   3. Resonance condition: $f_c = f_o$

   $\frac{3v}{4L_c} = \frac{2v}{L_o} \implies 3L_o = 8L_c \implies \frac{L_c}{L_o} = \frac{3}{8}$

   4. So $L_c : L_o = 3:8$.

   Common Mistakes

   • Using the closed pipe's harmonic sequence as if it supports all integers (using $n=3$ as if it were the 3rd harmonic of an "all harmonics" series) — closed pipes skip even harmonics, but $n=3$ itself is still a valid (odd) harmonic here, so this particular case works out, though the general rule must be kept in mind.
   • Mixing up the factor of 4 (closed pipe) vs 2 (open pipe) in the fundamental length-frequency relations.

   ✓Final answer

   The correct option is (B) — 3 : 8.

   ANSWER: B

4. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.A metal rod of 50 cm length is clamped at its midpoint and is set to vibrations. The density of that metal is $2 \times 10^2\ \mathrm{kg\,m^{-3}}$. Young's modulus of that metal is $8 \times 10^8\ \mathrm{N\,m^{-2}}$. Fundamental frequency of the vibration is. (A) 2500 Hz (B) 2 kHz (C) 2.75 kHz (D) 200 Hz

   ›Reveal solutionSolution

   A rod clamped at its middle vibrates with a node at the centre and antinodes at both ends; its fundamental frequency is $f=v/(2L)$ with $v=\sqrt{Y/\rho}$. Answer: (B) 2 kHz.

   Concept and Intuition

   Longitudinal standing waves in a rod are governed by the same wave equation as sound, with speed $v=\sqrt{Y/\rho}$ ($Y$= Young's modulus, $\rho$= density). Clamping the rod at its midpoint forces a node (zero displacement) there, while the two free ends must be antinodes (maximum displacement) — this is exactly analogous to a rod free at both ends but forced through the middle. The distance from the centre to each free end is a quarter wavelength ($\lambda/4$) in the fundamental mode, so the whole rod length $L$ corresponds to $\lambda/2$ ... equivalently the full length matches half of a full standing wave pattern that repeats over $2L$ — giving the standard textbook result $f=v/(2L)$ for this configuration.

   Step-by-Step Solution

   1. Compute the longitudinal wave speed in the rod: $v=\sqrt{\dfrac{Y}{\rho}}=\sqrt{\dfrac{8\times10^{8}}{2\times10^{2}}}=\sqrt{4\times10^{6}}=2000$ m/s.
   2. Rod length $L=50$ cm $=0.5$ m.
   3. Fundamental frequency for a rod clamped at the middle: $f=\dfrac{v}{2L}=\dfrac{2000}{2(0.5)}=\dfrac{2000}{1}=2000$ Hz.
   4. Convert: $2000\text{ Hz}=2\text{ kHz}$.

   Common Mistakes

   • Using $f=v/(4L)$ (the formula for a rod fixed at one end only) instead of $f=v/(2L)$ (fixed at the middle, free at both ends).
   • Forgetting to convert the rod length from cm to metres before substituting.

   ✓Final answer

   The correct option is (B) — 2 kHz.

   ANSWER: B

5. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A pipe with 30 cm Length is open at both ends. Which harmonic mode of the pipe resonates a 1.65 kHz source? (Velocity of sound in air $= 330\ \mathrm{ms^{-1}}$) (A) $2$ (B) $3$ (C) $3.5$ (D) $2.5$

   ›Reveal solutionSolution

   The fundamental frequency of the 30 cm open pipe is 550 Hz; $1650\ \text{Hz} = 3\times550\ \text{Hz}$ is exactly its third harmonic.

   Concept and Intuition

   A pipe open at both ends supports standing waves with antinodes at both open ends, allowing all harmonics (unlike a pipe closed at one end, which allows only odd harmonics). The resonant frequencies are integer multiples of the fundamental $\dfrac{v}{2L}$.

   Step-by-Step Solution

   1. Fundamental frequency: $f_1 = \dfrac{v}{2L} = \dfrac{330}{2\times0.30}=\dfrac{330}{0.6}=550\ \text{Hz}$.
   2. General harmonic: $f_n = n f_1 = 550n\ \text{Hz}$.
   3. Set equal to the source: $550n = 1650 \implies n = 3$.
   4. So the pipe resonates in its third harmonic ($n=3$) with this source.

   Common Mistakes

   • Using the closed-pipe formula (odd harmonics only, $f_n=(2n-1)v/4L$) instead of the open-pipe formula — the question explicitly states the pipe is open at both ends.
   • Forgetting to convert length to metres before dividing.

   ✓Final answer

   The correct option is (B) — $3$.

   ANSWER: B

6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The fundamental frequency of an open pipe is 100 Hz. If the bottom end of the pipe is closed and $\dfrac{1}{3}$rd of the pipe is filled with water, then the fundamental frequency of the pipe is (A) 200 Hz (B) 100 Hz (C) 75 Hz (D) 150 Hz

   ›Reveal solutionSolution

   Filling part of an open pipe with water turns it into a shorter closed pipe; recompute the fundamental frequency for the new closed-pipe length.

   Concept and Intuition

   An open pipe of length $L$ has both ends free to vibrate (antinodes at both ends), giving a fundamental $f=v/2L$. Once the bottom is sealed and water fills the lower third, the water surface behaves as a rigid, closed boundary (a node), while the top remains open (an antinode). The air column relevant to the vibration is now only the top two-thirds of the pipe, and it behaves exactly like a closed pipe of that shorter length, whose fundamental is $f=v/4L'$ (one quarter wavelength fits in the tube).

   Step-by-Step Solution

   1. Original open pipe: $f_0 = \dfrac{v}{2L} = 100\ \text{Hz} \Rightarrow \dfrac{v}{L}=200$.
   2. After filling: water occupies the bottom $\frac13$ of length $L$; the remaining air column length is $L' = \frac23 L$, closed at the water surface and open at the top.
   3. Fundamental of a closed pipe of length $L'$: $f' = \dfrac{v}{4L'} = \dfrac{v}{4\times\frac23L} = \dfrac{3v}{8L}$.
   4. Substitute $v/L = 200$: $f' = \dfrac{3\times200}{8} = \dfrac{600}{8} = 75$ Hz.

   Common Mistakes

   • Using the full length $L$ instead of the remaining $\frac23L$ air column.
   • Treating the new configuration as still an open pipe rather than recognizing the water surface creates a closed end.

   ✓Final answer

   The correct option is (C) — 75 Hz.

   ANSWER: C

7. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The vibrations of four air columns are shown below. The ratio of frequencies is [FIGURE] (four sub-diagrams labelled (a)-(d), each of length L, showing standing-wave loop patterns of vibrating air columns with different numbers of loops/nodes) (A) 1:2:3:4 (B) 1:3:2:4 (C) 1:4:3:2 (D) 1:4:2:3

   ›Reveal solutionSolution

   Each drawn air-column pattern corresponds to a specific closed-open or open-open pipe harmonic; reading off nodes/antinodes gives frequencies in the ratio 1:4:2:3.

   Concept and Intuition

   Standing waves in an air column must have a displacement NODE at any closed end and a displacement ANTINODE at any open end. The distance between successive nodes (or antinodes) is $\lambda/2$, and a full 'loop' pattern from node-to-node or antinode-to-antinode always spans $\lambda/2$. Reading the boundary conditions and the number of internal nodes/antinodes off a drawn vibration pattern directly tells us how many quarter- or half-wavelengths fit in the fixed length $L$, and hence the frequency (since $f=v/\lambda$).

   Step-by-Step Solution

   1. (a): starts as a single point at the left edge (a node — closed end) and flares out to touch both top and bottom at the right edge (an antinode — open end), with no other nodes in between. This is exactly ONE quarter-wavelength: $L=\lambda_a/4 \Rightarrow \lambda_a = 4L$, so $f_a = v/4L$.
   2. (b): touches both left and right edges fully (antinodes at both ends, open–open pipe) and has two internal crossing points (nodes) at $L/4$ and $3L/4$, i.e. antinodes at $0, L/2, L$ — three antinodes, two loops. This is the open-open pipe's 2nd harmonic: $L=\lambda_b \Rightarrow f_b = v/L = 4\left(\dfrac{v}{4L}\right) = 4f_a$.
   3. (c): touches both edges (antinodes at both ends — open-open) with a SINGLE crossing (node) exactly at the centre — this is the open-open pipe's fundamental (one full loop, half wavelength across $L$): $L=\lambda_c/2 \Rightarrow f_c = v/2L = 2f_a$.
   4. (d): shows the left portion closing into a node before the right edge, and the right portion a further node–antinode segment reaching the right edge — overall a node–antinode–node–antinode pattern, i.e. THREE quarter-wave segments in $L$ with a closed left end and open right end. This is the closed-open pipe's first overtone (3rd harmonic): $L=3\lambda_d/4 \Rightarrow f_d = 3v/4L = 3f_a$.
   5. Ratio: $f_a:f_b:f_c:f_d = 1:4:2:3$.

   Common Mistakes

   • Assuming all four patterns belong to the same type of pipe (all open-open or all closed-open) instead of reading the actual boundary condition (node vs antinode) drawn at each end.
   • Miscounting the number of loops/nodes in the more complex hourglass and mixed patterns (b) and (d).

   ✓Final answer

   The correct option is (D) — 1:4:2:3.

   ANSWER: D

8. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.The frequency of fifth harmonic of a closed pipe is equal to the frequency of third harmonic of an open pipe. If the length of the open pipe is 72 cm, then the length of the closed pipe is (A) 60 cm (B) 45 cm (C) 30 cm (D) 75 cm

   ›Reveal solutionSolution

   Equating the closed pipe's 5th-harmonic frequency to the open pipe's 3rd-harmonic frequency and solving for the closed pipe's length gives 60 cm.

   Concept and Intuition

   A closed pipe supports only odd harmonics with frequency $f_n=nv/4L$ ($n=1,3,5,\dots$), while an open pipe supports all harmonics with $f_n=nv/2L$. Setting the two given harmonic frequencies equal (same speed of sound $v$ in both) lets us solve directly for the unknown length.

   Step-by-Step Solution

   1. Closed pipe, 5th harmonic: $f=\dfrac{5v}{4L_c}$.
   2. Open pipe, 3rd harmonic: $f=\dfrac{3v}{2L_o}$, with $L_o=72$ cm.
   3. Equate: $\dfrac{5v}{4L_c}=\dfrac{3v}{2\times72}=\dfrac{3v}{144}$.
   4. Cross-multiply: $5\times144=4L_c\times3\Rightarrow720=12L_c\Rightarrow L_c=60$ cm.

   Common Mistakes

   • Using $4L$ for the open pipe or $2L$ for the closed pipe (swapping the fundamental relations).
   • Forgetting that a closed pipe only has odd harmonics (though the 5th here is legitimately odd, so no issue in this specific case, it's worth checking).

   ✓Final answer

   The correct option is (A) — 60 cm.

   ANSWER: A

9. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
   Q.A pipe of length 80 cm is open at both the ends. A second pipe, closed at one end, has the same fundamental frequency as the open pipe. The length of the second pipe is (A) 80 cm (B) 160 cm (C) 40 cm (D) 120 cm

   ›Reveal solutionSolution

   Matching the fundamental frequencies of an open pipe and a closed pipe shows the closed pipe must be exactly half as long.

   Concept and Intuition

   An open-open pipe of length $L$ supports a fundamental with wavelength $2L$, so $f_{\text{open}}=v/2L$. A pipe closed at one end supports a fundamental with wavelength $4L'$ (a quarter-wave resonance), so $f_{\text{closed}}=v/4L'$. Setting the two frequencies equal (same fundamental) directly compares $L$ and $L'$.

   Step-by-Step Solution

   1. $f_{\text{open}}=\dfrac{v}{2(80)}=\dfrac{v}{160}$.
   2. $f_{\text{closed}}=\dfrac{v}{4L'}$.
   3. Equate: $\dfrac{v}{160}=\dfrac{v}{4L'}\Rightarrow 4L'=160\Rightarrow L'=40$ cm.

   Common Mistakes

   • Using $4L$ for the open pipe or $2L$ for the closed pipe (swapping the harmonic relations).

   ✓Final answer

   The correct option is (C) — 40 cm.

   ANSWER: C

10. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.A music system playing a loud music is kept on a table. If a glass of water is kept on the same table, the glass vibrates and generates ripples in the water. This is due to a property known as (A) Reflection of the sound (B) Refraction of the sound (C) Reverberation of the sound (D) Resonance of the sound

    ›Reveal solutionSolution

    A glass on a table vibrating and rippling water when loud music plays nearby is a textbook case of resonance — the sound wave's frequency matching a natural vibration mode of the glass. Answer: Resonance.

    Concept and Intuition

    Every object has natural frequencies at which it vibrates most easily. When an external periodic driving force (here, sound pressure waves from the speaker, transmitted through the table) has a frequency at or near one of these natural frequencies, the object absorbs energy efficiently and its vibration amplitude grows dramatically. This amplitude build-up under a matching driving frequency is called resonance, and it's why the glass visibly shakes and ripples the water even though it isn't touched.

    Step-by-Step Solution

    1. The sound from the music system is a mechanical (pressure) wave that also propagates through the table into the glass.
    2. Reflection would just mean the sound wave bounces off a surface — it doesn't explain amplitude buildup in the glass itself.
    3. Refraction concerns bending of sound crossing between media — irrelevant here.
    4. Reverberation is about persistence of sound in a room due to multiple reflections — not about a glass shaking.
    5. Only when the driving frequency of the music coincides with (or is close to) the glass-water system's natural frequency does the glass vibrate strongly enough to visibly ripple the water — this is resonance.

    Common Mistakes

    • Confusing resonance with simple reflection or general "sound travels through solids" reasoning.
    • Mixing up reverberation (persistence/echo in a room) with resonance (frequency matching causing large-amplitude vibration).

    ✓Final answer

    The correct option is (D) — Resonance of the sound.

    ANSWER: D

11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.An open air pipe of length 80 cm has the second harmonic frequency equal to the fundamental frequency of a closed organ air pipe. The length of the closed pipe is (A) 20 cm (B) 40 cm (C) 60 cm (D) 10 cm

    ›Reveal solutionSolution

    Matches an open pipe's second harmonic to a closed pipe's fundamental; solving gives the closed pipe length as 20 cm.

    Concept and Intuition

    An open (both ends open) pipe supports all harmonics, $f_n=\dfrac{nv}{2L}$, while a closed (one end closed) pipe supports only odd harmonics of its fundamental, $f_n=\dfrac{(2n-1)v}{4L}$, with fundamental $f_1'=\dfrac{v}{4L'}$. Matching a harmonic of one pipe to a harmonic of the other lets us solve for an unknown length, with the speed of sound $v$ cancelling out.

    Step-by-Step Solution

    1. Open pipe, length $L=80\,\mathrm{cm}=0.8\,\mathrm{m}$: second harmonic $f_2 = \dfrac{2v}{2L} = \dfrac{v}{L} = \dfrac{v}{0.8}$.
    2. Closed pipe, unknown length $L'$: fundamental $f_1' = \dfrac{v}{4L'}$.
    3. Given $f_2 = f_1'$: $\dfrac{v}{0.8} = \dfrac{v}{4L'}$.
    4. Cancel $v$: $4L' = 0.8 \Rightarrow L' = 0.2\,\mathrm{m} = 20\,\mathrm{cm}$.

    Common Mistakes

    • Using $f_n=\dfrac{nv}{4L}$ for the open pipe (that formula is for the closed pipe only).
    • Forgetting the closed pipe's fundamental uses $4L'$ in the denominator, not $2L'$.

    ✓Final answer

    The correct option is (A) — 20 cm.

    ANSWER: A

12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1:2:3$. Then, the length of the segments are in the ratio ________ (A) $6:3:2$ (B) $4:3:2$ (C) $4:2:1$ (D) $3:2:1$

    ›Reveal solutionSolution

    Fundamental frequency is inversely proportional to segment length (same string, same tension), so frequencies $1:2:3$ correspond to lengths $6:3:2$.

    Concept and Intuition

    A vibrating string segment fixed at both ends has fundamental frequency $f = \dfrac{v}{2L}$, where $v=\sqrt{T/\mu}$ is the wave speed determined by the tension $T$ and linear mass density $\mu$. Since all three segments come from the SAME string (same tension throughout, same material/density), $v$ is identical for all three — so frequency depends only on length, and $f \propto 1/L$.

    Step-by-Step Solution

    1. $f \propto \dfrac{1}{L}$ for fixed $v$, so $L \propto \dfrac{1}{f}$.
    2. Given $f_1:f_2:f_3 = 1:2:3$, the lengths satisfy $L_1:L_2:L_3 = \dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}$.
    3. Clear the fractions by multiplying each term by the LCM of the denominators (LCM of 1, 2, 3 = 6): $L_1:L_2:L_3 = 6:3:2$.

    Common Mistakes

    • Directly taking the length ratio as $1:2:3$ (the same as the frequency ratio) — forgetting that frequency and length are inversely, not directly, related.

    ✓Final answer

    The correct option is (A) — $6:3:2$.

    ANSWER: A