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Sound waves and their properties Questions

Chapter: Oscillations and Waves | Total Available: 23 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

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 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

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

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 3 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

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

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 4 of 10 Authentic UPSC NDA Question

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

Step-by-Step Logic & Solution

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

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).

Question 5 of 10 Authentic UPSC NDA Question

Reverberation is a phenomenon associated with a

Step-by-Step Logic & Solution

Correct Answer

✓ Option B is correct. Reverberation is caused by multiple reflections of sound.

📖 Governing Law / Core Concept

Reverberation relates to the behavior of sound waves in enclosed spaces, governed by the principles of sound reflection.

🧪 Step-by-Step Breakdown

Reverberation occurs when sound waves reflect off surfaces multiple times before they reach the listener's ears, extending the duration of sound perception. This is quantitatively described by the reverberation time, \( T_{60} \), defined as the time it takes for sound to decay by 60 dB.

\[ T_{60} = 0.161 \times \frac{V}{A} \]

Where:

V: Volume of the room (cubic meters, m³)
A: Total surface area of the room (square meters, m²)

This formula illustrates that reverberation time is shorter in rooms with more surface area (absorptive materials) and larger volumes.

🔍 Option Analysis

Option A refers to refraction, which involves the bending of sound waves; Option C implies a single refraction, which doesn't apply; Option D mentions single reflection, inadequate for sound persistence. None of these describe reverberation.

⚡ Mnemonic / Speed-Run

Remember: "Reverberation Rings" - Multiple reflections create continuous sound.

6. Visual Suggestion

A schematic diagram showing sound waves bouncing off walls of a concert hall, visualizing how sound reflects multiple times, indicating increased reverberation time.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter on Sound.

Question 6 of 10 Authentic UPSC NDA Question

Which among the following is true for propagation of sound waves?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (b) is correct. Sound cannot travel in vacuum and it is a longitudinal wave in air.

📖 Governing Law / Core Concept

Sound waves are mechanical waves governed by the principle that they require a medium (solid, liquid, or gas) to propagate. They are classified as longitudinal waves because the particle displacement is parallel to the wave direction.

🧪 Step-by-Step Breakdown

1. Mechanical nature: Sound requires a medium, thus cannot travel through vacuum.

2. Wave type: In sound propagation, particles in the medium vibrate back and forth in the same direction as the wave travels, indicating a longitudinal wave motion.

Mathematically, the speed of sound in a gas can be expressed as:

\[ v = \sqrt{\frac{B}{\rho}} \]

where:

v = speed of sound (m/s)
B = bulk modulus of the medium (N/m²)
\(\rho\) = density of the medium (kg/m³)

🔍 Option Analysis

A: Incorrect. Sound cannot travel in a vacuum.
C: Incorrect. Sound does not travel in vacuum.
D: Incorrect. Sound is not a transverse wave.

⚡ Mnemonic / Speed-Run

Remember: "Sound Needs a Medium" to recall that sound cannot travel through vacuum.

6. Visual Suggestion

Illustrate a longitudinal wave with particle motion depicted in a spring or a slinky to show compression and rarefaction, highlighting that particle displacement is parallel to wave direction.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 7 of 10 Authentic UPSC NDA Question

At 20° C, the speed of sound in water is approximately

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. The speed of sound in water at 20° C is approximately 1500 m/s.

📖 Governing Law / Core Concept

The speed of sound in any medium is influenced by its density and elasticity. In water, the speed of sound is approximately given by the formula:

\[ v = \sqrt{\frac{E}{\rho}} \]

where \( v \) is the speed of sound, \( E \) is the modulus of elasticity, and \( \rho \) is the density of the medium.

🧪 Step-by-Step Breakdown

At 20° C, the following values apply to water:

Density \( \rho \approx 1000 \, \text{kg/m}^3 \)
Bulk modulus (approximate elasticity) \( E \approx 2.2 \times 10^9 \, \text{N/m}^2 \)

Substituting these values into the formula:

\[ v = \sqrt{\frac{2.2 \times 10^9}{1000}} \]

\[ v = \sqrt{2.2 \times 10^6} \approx 1483 \, \text{m/s} \]

This approximation rounds to approximately 1500 m/s, affirming option (c).

🔍 Option Analysis

Option (a) 330 m/s: Typical speed of sound in air at room temperature.
Option (b) 800 m/s: Incorrect; represents a low estimate.
Option (d) 5000 m/s: Exceeds normal values for any non-metallic medium.

⚡ Mnemonic / Speed-Run

Remember, the speed of sound in water is around 1500 m/s; think of it as swift as a swimmer!

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against standard physics texts and research on sound propagation in fluids.

Question 8 of 10 Authentic UPSC NDA Question

The loudness of sound depends upon the

Step-by-Step Logic & Solution

Correct Answer

✓ Option (b) is correct. Loudness is determined by the amplitude of sound waves.

📖 Governing Law / Core Concept

The loudness of sound relates to its amplitude, represented in physics as:

\[ \text{Loudness} \propto A^2 \]

Where \( A \) is the amplitude of the sound wave.

🧪 Step-by-Step Breakdown

Loudness is a subjective perception of sound intensity. It directly depends on the amplitude of the sound waves, which dictates how much energy is carried by the wave. Greater amplitude results in louder sounds.

🔍 Option Analysis

A: Velocity of sound waves affects how sound travels but not its loudness.
C: Frequency affects pitch, not loudness.
D: Frequency and velocity together do not contribute to loudness; only amplitude does.

⚡ Mnemonic / Speed-Run

Remember: "Amplitude Affects Loudness!" This simple phrase helps recall that louder sounds come from greater wave amplitudes.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 12 (Sound).

Question 9 of 10 Authentic UPSC NDA Question

Compared to audible sound waves, ultrasound waves have

Step-by-Step Logic & Solution

Correct Answer

✓ Option (b) is correct. Ultrasound waves possess a higher frequency than audible sound waves.

📖 Governing Law / Core Concept

Ultrasound waves are defined as sound waves with frequencies above the audible range, typically greater than 20 kHz.

🧪 Step-by-Step Breakdown

The relationship between frequency (f), speed (v), and wavelength (λ) of a wave is given by:

\[ v = f \cdot \lambda \]

For audible sound waves, frequency ranges from approximately 20 Hz to 20 kHz.
For ultrasound, frequencies exceed 20 kHz, indicating higher f.

🔍 Option Analysis

A: Higher speed. Incorrect, as speed is generally similar in both media (air, water).
C: Longer wavelength. Incorrect; higher frequency corresponds to shorter wavelengths.
D: Both higher speed and frequency. Incorrect, only frequency is higher without speed difference in same medium.

⚡ Mnemonic / Speed-Run

Remember: "High Frequency, Low Wavelength" for understanding ultrasound.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

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

Question 10 of 10 Authentic UPSC NDA Question

Which of the following statements about sound waves is correct? 1. Sound waves can travel through a vacuum. 2. Sound waves are longitudinal waves. 3. The speed of sound is faster in air than in water.

Step-by-Step Logic & Solution

[CORE CONCEPT]: Sound waves are mechanical waves that require a medium to travel and are classified as longitudinal waves. [STEP-BY-STEP SOLUTION]: Statement 1 is incorrect because sound cannot travel through a vacuum. Statement 2 is correct. Statement 3 is incorrect; sound travels faster in water than in air. [SHORTCUT / EXAM TRICK]: Remember that sound needs a medium (solid, liquid, gas) to propagate. [COMMON MISTAKE TRAP]: Confusing the properties of sound waves with electromagnetic waves, which can travel through a vacuum.