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Sound waves and their properties

Chapter: Oscillations and Waves

Question 1 of 5 NDA MCQ

A sound wave has a frequency of 1 kHz andwavelength 50 cm. How long will it take totravel 1 km?

Detailed Explanation

Correct Answer

✓ Option (d) is correct. The sound wave will take 2 seconds to travel 1 km.

📖 Governing Law / Core Concept

The speed of sound is defined by the equation:
\[ v = f \lambda \]
where:
  • v: Speed of sound (meters per second, m/s)
  • f: Frequency (hertz, Hz)
  • \(\lambda\): Wavelength (meters, m)

🧪 Step-by-Step Breakdown

Given:
  • Frequency \( f = 1 \text{ kHz} = 1000 \text{ Hz} \)
  • Wavelength \( \lambda = 50 \text{ cm} = 0.5 \text{ m} \)
Calculating speed:
\[ v = 1000 \text{ Hz} \times 0.5 \text{ m} = 500 \text{ m/s} \]
To find the time \( t \) taken to travel 1 km (1000 m): \div style="text-align: center; margin: 10px 0; overflow-x: auto; -webkit-overflow-scrolling: touch; max-width: 100%; padding: 5px 0;">\[ t = \frac{d}{v} = \frac{1000 \text{ m}}{500 \text{ m/s}} = 2 \text{ s} \]

📌 Hints / Properties Used:

  • Speed of Sound Equation: \( v = f \lambda \)
  • Distance-Time-Speed Relation: \( t = \frac{d}{v} \)

📊 Visual Diagram Suggestion:

Illustration depicting the sound wave propagation through a medium, with measurements indicating frequency, wavelength, and the distance traveled (1 km). Use arrows to denote direction and speed to enhance clarity.

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Waves).

Question 2 of 5 NDA MCQ

The frequency (f), wavelength (lambda) and speed(v) of a sound wave are related as

Detailed Explanation
✓

1. Correct Answer

Option D is correct. [The speed of sound is defined by the equation v = fλ.]

📖 Core Concept

The speed of sound in any medium is determined by its frequency and wavelength, expressed as v = fλ.

💡 3. Explanation

The relationship between frequency \( f \), wavelength \( \lambda \), and speed \( v \) is defined as:
v = fλ
Therefore, the speed of sound is directly proportional to its frequency and wavelength.

❌ 4. Option Analysis

  • •Option A: Incorrect, it asserts f = vλ, which reverses the relation.
  • •Option B: Incorrect as it states λ = vf, a misrepresentation of the relationship.
  • •Option C: Incorrect, it lacks a comprehensible physical connection.
⚡

NDA Speed-Run / Mnemonic Shortcut

Memorize v = fλ to quickly apply the relationship for sound waves.

🎯

5. Key Takeaways

The speed of sound is given by the relation v = fλ.

Question 3 of 5 NDA MCQ

The sound created in a big hall persists because of the repeated reflections. The phenomenon is called

Detailed Explanation

Correct Answer

✓ Option (a) is correct. The phenomenon is known as reverberation.

📖 Governing Law / Core Concept

Reverberation occurs due to the repeated reflection of sound waves in large spaces. The basic principle is formulated as: \[ \text{Sound Intensity} \propto \frac{1}{d^2} \] where \( d \) is the distance from the sound source.

🧪 Step-by-Step Breakdown

When sound waves encounter surfaces, they are reflected back multiple times before they dissipate. This results in the sound persisting in the environment. The reverberation time (\( T_r \)) can be calculated using Sabine's formula: \[ T_r = \frac{0.161 V}{A} \] where: - \( V \) = volume of the room (m³) - \( A \) = total surface area of absorption (m²) This reflection creates a layered effect of sound that can enhance auditory experiences, typical in concert halls.

🔍 Option Analysis

- **B: Dispersion** refers to the separation of waves into their component frequencies, not applicable here. - **C: Refraction** is the bending of waves as they pass through different media, irrelevant to sound persistence. - **D: Diffraction** is the bending of waves around obstacles; it contributes to sound behavior but not to persistent sound in a hall.

⚡ Mnemonic / Speed-Run

To remember reverberation, think "Room Echo" — R for Room, E for Echo.

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating sound wave propagation and reflection within a hall, showing how sound reflects off walls creating multiple sound paths leading to a persistent auditory experience.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter 15 (Waves).

Question 4 of 5 NDA MCQ

Which one of the following cannot be the unit of frequency of a sound wave?

Detailed Explanation

Correct Answer

✓ Option (a) is correct. dB (decibel) is not a unit of frequency.

📖 Governing Law / Core Concept

The frequency of a sound wave is defined as the number of oscillations or cycles it completes per unit time. It is measured in Hertz (Hz), where \(1\, \text{Hz} = 1\, \text{s}^{-1}\).

🧪 Step-by-Step Breakdown

The unit of frequency can be defined as:
\[ \text{Frequency} \, (f) = \frac{\text{Number of cycles}}{\text{Time interval}} \]
This leads us to the units: - Hertz (Hz) or cycles per second: \( \text{Hz} = \text{s}^{-1} \) - Minutes inverse (\( \text{min}^{-1} \)) can also be used, as it is simply cycles per minute. However, the decibel (dB) is a logarithmic unit used for ratios of power or intensity, not frequency.

🔍 Option Analysis

- **A: dB** - Decibel, a unit for measuring sound intensity, not frequency. - **B: s⁻¹** - Equivalent to Hertz, valid unit of frequency. - **C: Hz** - Standard unit of frequency (1 Hz = 1 cycle/s). - **D: min⁻¹** - Valid unit representing frequency (cycles per minute).

⚡ Mnemonic / Speed-Run

Remember: "Hertz for cycles, seconds for time; anything else is off its prime."

6. Visual Suggestion

6. Visual Suggestion

Visualize a sound wave oscillating through time with its frequency labeled in Hz, dB, and min⁻¹. Illustrate a wave's upward and downward cycles within one second to relate distance between cycles to frequency.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Sound).

Question 5 of 5 NDA MCQ

Beats is a phenomenon that occurs when frequencies of two harmonic waves are

Detailed Explanation

Correct Answer

✓ Option (d) is correct. Beats occur due to two waves having nearly the same frequencies.

📖 Governing Law / Core Concept

Beats are a phenomenon resulting from the interference of two harmonic waves with closely matched frequencies, represented mathematically by:

\[ f_{\text{beats}} = |f_1 - f_2| \]

where \( f_1 \) and \( f_2 \) are the frequencies of the two waves (in Hz). SI unit is Hertz (Hz).

🧪 Step-by-Step Breakdown

When two waves \( y_1 = A \sin(2 \pi f_1 t) \) and \( y_2 = A \sin(2 \pi f_2 t) \) interfere, the resultant wave is given by:

\[ y = y_1 + y_2 = A \sin(2 \pi f_1 t) + A \sin(2 \pi f_2 t) \]

Using the sine addition formula, this can be simplified to observe beat frequencies. When \( f_1 \) and \( f_2 \) are nearly the same, the interference results in alternating constructive and destructive wave patterns, leading to beats.

🔍 Option Analysis

Option (a): Equal frequencies lead to a steady wave, not beats. Option (b): Far apart frequencies produce no beats, only distinct waves. Option (c): Multiples lead to harmonics, not beats.

⚡ Mnemonic / Speed-Run

Remember: "Beats Occur when Frequencies are Close (BOFC)" to quickly identify the conditions for beat formation.

Visual Suggestion

Visual Suggestion

A wave diagram showing two waves with slightly different frequencies (e.g., 440 Hz and 442 Hz) illustrating how their superposition results in a beat frequency visible in amplitude variations.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 15 (Waves).

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