A bat emits ultrasonic sound of frequency $1000\ \text{kHz}$ in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is $340\ \text{m s}^{-1}$ and in water $1486\ \text{m s}^{-1}$.

A bat emits ultrasonic sound of frequency 1000kHz in air. If the sound meets a water surface, what is the wave

Exercises · 14.6

1000\ \text{kHz}

(a) the reflected sound,

(b) the transmitted sound? Speed of sound in air is

340\ \text{m s}^{-1}

and in water

1486\ \text{m s}^{-1}

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Concept understanding — Acoustic Resonance Harmonics

Acoustic Resonance Harmonics

Imagine pushing a child on a swing. If you push at random moments, the swing jerks but never goes high. But if you push exactly when the swing is coming back toward you — matching its natural rhythm — each small push adds to the motion, and soon the swing soars. That is resonance: a small, well-timed force builds up a large response.

Acoustic resonance is the same idea, but with sound. A guitar string, an air column in a pipe, or a wine glass each has certain natural frequencies at which it vibrates easily. When a sound wave (or a periodic push) arrives at one of those frequencies, the object absorbs energy efficiently and its vibration amplitude grows large. That build-up is acoustic resonance.

Harmonics: The Family of Natural Frequencies

An object does not have just one natural frequency — it has a whole ladder of them, called harmonics. The lowest one is the fundamental (first harmonic); the rest are related to it in a way that depends on the boundary conditions of the vibrating system. This is the point most notes skip, and it is exactly what CBSE Class 11 tests.

Important

The statement "harmonic frequencies are whole-number multiples of the fundamental" is only true for systems that are symmetric at both ends (both fixed, or both free). It is not true for every vibrating system — a stretched drum membrane, for instance, has overtones that are not simple whole-number multiples of its fundamental, which is exactly why a drum's note sounds less "musical" than a string's.

Case 1 — Both ends fixed (a stretched string) or both ends open (an open organ pipe)

Here every harmonic is present:

f_n = n f_1, \qquad n = 1, 2, 3, 4, \dots

For a string fixed at both ends, the fundamental $f_1$ is one loop; $f_2 = 2f_1$ is two loops, $f_3 = 3f_1$ is three loops, and so on. A pipe open at both ends behaves the same way for the air column inside it.

Case 2 — One end closed, one end open (a closed organ pipe)

The closed end must be a displacement node and the open end an antinode. That boundary condition rules out the even harmonics — only the odd multiples of the fundamental survive:

f_n = n f_1, \qquad n = 1, 3, 5, 7, \dots

This is why a closed pipe of a given length sounds an octave lower (and tonally different) than an open pipe of the same length — it is missing every even harmonic.

The Precise Statement

Acoustic resonance harmonics occur when a driving sound wave's frequency matches one of the natural frequencies of a vibrating system, causing that system to vibrate with maximum amplitude at that natural frequency. Which harmonics exist — all integers, or only odd integers — depends entirely on the boundary conditions at the two ends of the system.

Why This Matters

A guitar string vibrates at its fundamental and several of its harmonics simultaneously; the relative strength of each harmonic gives the instrument its characteristic timbre. …

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When sound crosses a boundary, frequency stays constant but wavelength changes with the medium's wave speed. Reflected sound: $\lambda = 3.4 \times 10^{-4}\ \text{m}$ ; transmitted sound: $\lambda = 1.486 \times 10^{-3}\ \text{m}$ .

Why frequency is conserved at a boundary

When a wave encounters the interface between two media, the boundary itself oscillates at a single frequency — it cannot vibrate at two different rates simultaneously. This physical constraint forces the frequency of the reflected wave (staying in air) and the transmitted wave (entering water) to match the incident frequency exactly. What does change is the wavelength, because the wave speed differs in each medium.

The fundamental wave relation $v = f \lambda$ governs everything. Since $f$ is locked, wavelength must adjust: $\lambda = \frac{v}{f}$ .

Step-by-step solution

1. Identify the given data

Frequency of ultrasonic sound: $f = 1000\ \text{kHz} = 1000 \times 10^3\ \text{Hz} = 10^6\ \text{Hz}$
Speed of sound in air: $v_{\text{air}} = 340\ \text{m s}^{-1}$
Speed of sound in water: $v_{\text{water}} = 1486\ \text{m s}^{-1}$

2. Find the wavelength of the reflected sound

The reflected sound bounces back into the air, so it travels in the same medium as the incident wave. Both the speed and frequency remain unchanged.

Using $\lambda = \frac{v}{f}$ :

\lambda_{\text{reflected}} = \frac{v_{\text{air}}}{f} = \frac{340}{10^6} = 3.4 \times 10^{-4}\ \text{m} = 0.34\ \text{mm}

3. Find the wavelength of the transmitted sound

The transmitted sound enters water. The frequency is still $f = 10^6\ \text{Hz}$ (conserved at the boundary), but now the wave speed is $v_{\text{water}}$ .

$\lambda_{\text{transmitted}} = \frac{v_{\text{water}}}{f} = \frac{1486}{10^6} = 1.486 \times 10^{-3}\ \text{m} = 1.486\ \text{mm}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The frequency of the note emanating from an organ pipe depends on$
(a) length of the air column
(b) the speed of sound in air
(c) both (A) and (B)
(d) none of these
›Reveal solutionSolution
Organ-pipe frequency depends on both air-column length and sound speed. Answer (C).

For an air column, the fundamental frequency is:

open pipe: f = v/(2L)

closed pipe: f = v/(4L) …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. The ratio of frequencies of harmonics produced from closed organ pipe and open organ pipe of same length is (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1$
›Reveal solutionSolution
For the same length, the fundamental frequency of a closed pipe is half that of an open pipe — ratio 1:2.

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

Closed organ pipe (closed at one end): fundamental frequency $f_{closed} = \dfrac{v}{4L}$ …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Resonance tube is a/an (A) Closed organ pipe (B) Open organ pipe (C) Flow tube (D) Exit tube$
›Reveal solutionSolution

A resonance tube behaves as a closed organ pipe.

In the resonance tube experiment, a vertical tube is partly filled with water; the water surface acts as a rigid, closed boundary (a displacement node forms there) while the top of the tube is open to the atmosphere (a displacement antinode forms there). This is exactly the boundary condition of a pipe closed at one end and open at the other — i.e. a closed or …

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CBSE 2023Set ANNUAL1 mark
$Q. What will be the ratio of fundamental frequencies of an open and closed organ pipe of the same length?$
›Reveal solutionSolution

Comparing $f_{open} = v/2L$ with $f_{closed} = v/4L$ for the same length L gives a ratio of 2:1.

In an open organ pipe (open at both ends), antinodes form at both ends, and the fundamental mode fits half a wavelength into the pipe length $L$ : $L = \lambda_{open}/2$ , so $\lambda_{open} = 2L$ , giving fundamental frequency

$f_{open} = \frac{v}{\lambda_{open}} = \frac{v}{2L}$

In a closed organ pipe (closed at one end, open at the other), a node forms at the closed end and an antinode at the open end, and the fundamental mode fits a quarter wavelength into the length $L$ : $L = \lambda_{closed}/4$ , so $\lambda_{closed} = 4L$ , giving fundamental frequency …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The fundamental frequency of closed organ pipe whose length is 10 cm is :$
(a) 4.5 vHz
(b) 2.5 vHz
(c) 10 vHz
(d) 2 vHz
›Reveal solutionSolution

A closed organ pipe has a node at the closed end and an antinode at the open end, giving fundamental frequency f = v/(4L) = 2.5v Hz for L = 0.1 m.

A closed organ pipe (closed at one end, open at the other) vibrates with a node at the closed end and an antinode at the open end. The fundamental (first) mode fits a quarter wavelength inside the pipe length L:

L = lambda/4, so lambda = 4L

The fundamental frequency is

f = v/lambda = v/(4L)

Given L = 10 cm = 0.1 m,

f = v/(4 x 0.1) = v/0.4 = 2.5v Hz

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The first three frequencies of harmonics of a closed organ pipe will be in the ratio:$
(a) 1:2:3
(b) 1:3:5
(c) 1:4:9
(d) 2:4:6
›Reveal solutionSolution

A pipe closed at one end must have a node at the closed end and an antinode at the open end, which allows only odd-integer multiples of the fundamental frequency — the ratio 1:3:5.

In a closed organ pipe (closed at one end, open at the other), the closed end must always be a displacement node and the open end an antinode. This boundary condition permits only standing waves whose length corresponds to an odd number of quarter-wavelengths fitting the pipe length L:

L = (2n-1)λ/4, for n = 1, 2, 3, …

The allowed frequencies are then:

f_n = (2n-1)v/4L = (2n-1) f1

…

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CBSE 2018Set hz1 markMCQ
$Q. The harmonics present in an open organ pipe are: (A) Odd harmonics (B) Even harmonics (C) Even as well as odd harmonics (D) None of these$
›Reveal solutionSolution

An open organ pipe (open at both ends) has antinodes at both ends and supports all harmonics — both even and odd multiples of the fundamental frequency.

An organ pipe open at both ends has displacement antinodes at both open ends. The allowed modes of vibration must fit an integer number of half-wavelengths between the two ends:

L = n (lambda_n / 2), n = 1, 2, 3, ...

So the frequencies of the harmonics are:

f_n = n v / (2L) = n f_1

where f_1 = v/2L is the fundamental frequency.

This means the open pipe supports f_1, 2f_1, 3f_1, 4f_1, ... — i.e., ALL integer harmonics (both odd: 1st, 3rd, 5th... and even: 2nd, 4th, 6th...) are present.

(This is in contrast to a pipe closed at one end, which has a node at the closed end and an antinode at the open end, and supports only odd harmonics: f_1, 3f_1, 5f_1, ...)

Checking options: …

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Q.A bat emits ultrasonic sound of frequency 1000 kHz1000\ \text{kHz}1000 kHz in air. If the sound meets a water surface, what is the wavelength of

Why frequency is conserved at a boundary

Step-by-step solution

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Q.A bat emits ultrasonic sound of frequency $1000\ \text{kHz}$ in air. If the sound meets a water surface, what is the wavelength of

Unlock everything free for 14 days