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 (a) $\lambda$ (b) $\dfrac{\lambda}{2}$ (c) $2\lambda$ (d) $4\lambda$

Sound waves of wavelength λ travelling in a medium with a speed of v m/s enter into another medium where its s

NCERT Exemplar · Q2

\lambda

(a)

\lambda

(b)

\dfrac{\lambda}{2}

(c)

2\lambda

(d)

4\lambda

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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.
A clarinet (acoustically closer to a closed pipe) is rich in odd harmonics, giving it a distinctive "hollow" tone compared with a flute (acoustically an open pipe), which produces all harmonics.
A singer can shatter a wine glass by singing exactly at its resonant frequency — the glass absorbs energy from the sound wave until the vibration amplitude exceeds what the glass can withstand.

Watch out

Do not assume every vibrating object supports the full integer series $f_n = nf_1$ . Always check the boundary conditions first: symmetric ends (both fixed/both free/both open) give all harmonics; one fixed + one free (or closed + open) gives only odd harmonics; and two-dimensional systems like membranes do not follow a simple integer ladder at all.

The Intuition in One Sentence

Acoustic resonance harmonics are the "sweet spots" where a system vibrates most easily — a ladder of natural frequencies built on the fundamental, whose rungs (all integers, or odd integers only) are decided by how the two ends of the system are constrained.

This topic is commonly searched as "Acoustic Resonance Harmonics 11 physics important questions" or "Acoustic Resonance Harmonics formula and examples", and it maps cleanly onto the Class 11 Physics portion of the NCERT/CBSE syllabus. Because acoustic resonance harmonics shows up repeatedly in JEE Main, NEET and state engineering/medical entrance exams, mastering the underlying idea (not just the formula) is genuinely worth the extra time.

When a wave crosses into a new medium its frequency remains constant while speed and wavelength adjust together; doubling the speed doubles the wavelength to $2\lambda$ .

Why frequency is the bridge between media

When any wave—sound, light, water—crosses from one medium into another, one quantity stays absolutely fixed: frequency. Think of it this way: if the source vibrates 440 times per second (say, an A note), every crest that arrives at the boundary must continue into the second medium. The boundary cannot create or destroy oscillations; it can only transmit them. So the number of wave cycles passing any point per second—the frequency $f$ —is the same on both sides.

What does change? The wave speed $v$ , because the new medium has different physical properties (density, elasticity, etc.). And because speed, frequency, and wavelength are locked together by the fundamental wave relation

v = f \lambda,

the wavelength $\lambda$ must adjust to keep this equation balanced when $v$ changes but $f$ does not.

Step-by-step reasoning

1. Write the wave relation in the first medium.

In the original medium the wave travels at speed $v$ with wavelength $\lambda$ , so

v = f \lambda.

Solve for the frequency:

f = \frac{v}{\lambda}.

2. Recognize that frequency is unchanged in the second medium.

The wave enters the second medium with the same frequency $f = \frac{v}{\lambda}$ , because the source oscillation rate has not changed and the boundary transmits every cycle.

3. Write the wave relation in the second medium.

Now the speed is $2v$ and the wavelength is unknown; call it $\lambda'$ . The wave equation gives

2v = f \lambda'.

4. Substitute the frequency and solve for $\lambda'$ .

Replace $f$ with $\frac{v}{\lambda}$ :

2v = \frac{v}{\lambda} \cdot \lambda' \quad \Rightarrow \quad \lambda' = \frac{2v \cdot \lambda}{v} = 2\lambda.

The wavelength in the second medium is exactly twice the original.

Tip

A quick proportionality: $\lambda \propto v$ when $f$ is constant. If speed doubles, wavelength doubles; if speed halves, wavelength halves.

Watch out

Do not assume wavelength stays constant across media—that is true only for frequency. Wavelength and speed always change together to preserve $v = f\lambda$ .

✓Final answer

2\lambda

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Q.Sound waves of wavelength λ\lambdaλ travelling in a medium with a speed of vvv m/s enter into another medium where its speed is 2v2v2v m/s. Wavelength of sound waves in the second medium is

Why frequency is the bridge between media

Step-by-step reasoning

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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

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