3D UNIVERSAL ENGLISH INSITUTE INC
info.3duniversal.com@gmail.com
8:00-17:00(Mon-Fri)

Sound: NMAT Physics Review

Sound: NMAT Physics Review

Sound is a fundamental topic in physics and a recurring subject in the NMAT Physics section. It combines concepts from waves, mechanics, and human perception, making it both theoretical and practical. A solid understanding of sound involves knowing how sound waves are produced, how they propagate through different media, how their properties are measured, and how humans perceive them. This review is designed to comprehensively cover all essential sound concepts relevant to the NMAT, with clear explanations and exam-oriented emphasis.

Nature of Sound Waves

Sound is a form of mechanical wave that results from the vibration of particles in a medium. Unlike electromagnetic waves, sound cannot propagate through a vacuum because it requires a material medium—such as air, liquid, or solid—to travel. Sound waves are classified as longitudinal waves, meaning that the particle motion is parallel to the direction of wave propagation.

In a longitudinal sound wave, regions of high pressure (compressions) alternate with regions of low pressure (rarefactions). As the wave moves, particles oscillate back and forth about their equilibrium positions without permanently changing location. Energy is transferred through the medium via these oscillations.

Production of Sound

Sound is produced when an object vibrates. These vibrations cause nearby particles in the medium to oscillate, generating a sound wave. Common examples include:

  • Vocal cords vibrating to produce speech
  • A guitar string vibrating when plucked
  • A tuning fork vibrating when struck

The frequency of vibration determines the pitch of the sound, while the amplitude of vibration determines its loudness. The nature of the vibrating source largely defines the characteristics of the sound produced.

Propagation of Sound Waves

Sound waves propagate by transferring energy through particle interactions within a medium. The speed of sound depends on the physical properties of the medium, particularly its elasticity and density.

In general:

  • Sound travels fastest in solids
  • Slower in liquids
  • Slowest in gases

For example, the speed of sound in air at room temperature (approximately 20°C) is about 343 m/s. In water, it is roughly 1,500 m/s, and in steel, it can exceed 5,000 m/s.

Speed of Sound in Gases

In gases, the speed of sound is influenced mainly by temperature rather than pressure. The relationship between speed of sound and temperature in an ideal gas is given by:

v = √(γRT / M)

where:

  • v = speed of sound
  • γ = ratio of specific heats
  • R = universal gas constant
  • T = absolute temperature
  • M = molar mass of the gas

An increase in temperature increases the speed of sound because molecules move faster, allowing pressure disturbances to propagate more quickly.

Frequency, Wavelength, and Wave Speed

Sound waves obey the fundamental wave equation:

v = fλ

where:

  • v = wave speed
  • f = frequency
  • λ = wavelength

For a given medium, the speed of sound remains constant. Therefore, an increase in frequency results in a decrease in wavelength, and vice versa. This relationship is crucial for solving many NMAT problems involving sound waves.

Sound Intensity and Intensity Level

Sound intensity (I) is defined as the power transmitted per unit area perpendicular to the direction of wave propagation. It is measured in watts per square meter (W/m²). Intensity decreases with distance from the source due to the spreading of wave energy.

Because the human ear can detect a wide range of sound intensities, a logarithmic scale is used to describe sound loudness. This scale is known as the sound intensity level (β), measured in decibels (dB).

β = 10 log₁₀ (I / I₀)

where I₀ = 1 × 10⁻¹² W/m² (threshold of hearing).

Loudness and Human Perception

Loudness is a subjective sensation that depends on sound intensity as well as the sensitivity of the human ear. Two sounds with the same intensity may not be perceived as equally loud if they have different frequencies.

The human ear is most sensitive to frequencies between 2,000 Hz and 4,000 Hz. NMAT questions often test the distinction between physical intensity and perceived loudness.

Pitch and Frequency

Pitch is the auditory sensation corresponding to the frequency of a sound wave. Higher frequency sounds are perceived as higher pitch, while lower frequency sounds are perceived as lower pitch.

The audible frequency range for humans is approximately 20 Hz to 20,000 Hz. Frequencies below this range are called infrasonic, while those above are ultrasonic.

Quality or Timbre of Sound

Timbre refers to the quality of sound that allows us to distinguish between different sound sources producing the same pitch and loudness. It depends on the waveform of the sound and the presence of overtones or harmonics.

Musical instruments produce complex sounds composed of a fundamental frequency and multiple harmonics. The relative intensities of these harmonics determine the timbre.

Reflection of Sound and Echo

Sound waves can reflect off surfaces, similar to light waves. An echo occurs when reflected sound reaches the listener after a sufficient time delay (at least 0.1 seconds).

The minimum distance required to hear a distinct echo in air is approximately 17 meters, assuming a speed of sound of 340 m/s.

Echoes are used in applications such as sonar, echolocation by bats, and medical ultrasound imaging.

Refraction and Diffraction of Sound

Sound waves undergo refraction when they pass through regions of different temperatures or densities, causing a change in wave speed and direction. This phenomenon explains why sound can travel farther at night when temperature gradients are present.

Diffraction allows sound waves to bend around obstacles and spread through openings. This is why sound can be heard even when the source is not in direct line of sight.

Interference and Beats

When two sound waves of similar frequencies interfere, they produce beats. The beat frequency is equal to the absolute difference between the two frequencies:

fbeats = |f₁ − f₂|

Beats are commonly used to tune musical instruments and are frequently tested in NMAT problem-solving questions.

Doppler Effect

The Doppler Effect refers to the apparent change in frequency of a sound wave due to relative motion between the source and the observer. When the source approaches the observer, the observed frequency increases; when it moves away, the frequency decreases.

The Doppler Effect formula for sound is:

f’ = f (v ± vo) / (v ∓ vs)

where:

  • f’ = observed frequency
  • f = source frequency
  • v = speed of sound
  • vo = observer velocity
  • vs = source velocity

Standing Waves and Resonance

Standing waves are formed when two waves of the same frequency and amplitude travel in opposite directions. They are characterized by nodes (points of zero displacement) and antinodes (points of maximum displacement).

Resonance occurs when a system vibrates at its natural frequency with maximum amplitude. In sound, resonance enhances loudness and is fundamental to the operation of musical instruments.

Sound in Pipes

Sound waves in pipes form standing waves. There are two main types of pipes:

  • Open pipes (both ends open)
  • Closed pipes (one end closed)

Open pipes support all harmonics, while closed pipes support only odd harmonics. Understanding these patterns is essential for NMAT questions involving resonance and harmonics.

Applications of Sound

Sound has numerous practical applications, including:

  • Medical ultrasound imaging
  • Sonar for underwater navigation
  • Noise control and acoustics
  • Musical instrument design

NMAT questions may test conceptual understanding of these applications rather than detailed technical knowledge.

Key NMAT Exam Tips for Sound

  • Understand relationships between frequency, wavelength, and speed
  • Differentiate between physical quantities and human perception
  • Memorize key formulas and when to apply them
  • Practice Doppler Effect and beat frequency problems

Mastery of sound concepts strengthens overall performance in NMAT Physics. With consistent practice and conceptual clarity, sound-related questions can become a reliable scoring area in the exam.

Problem Set: Sound (NMAT Physics)

  1. Wave Speed Relation
    A sound wave in air has a frequency of 680 Hz and a wavelength of 0.50 m. What is the speed of sound in air?
  2. Wavelength from Frequency
    The speed of sound in air is 340 m/s. What is the wavelength of a 170 Hz tone?
  3. Echo Distance
    A person stands 85 m away from a large wall and shouts. How long will it take for the echo to return? (Assume speed of sound = 340 m/s.)
  4. Minimum Distance for an Echo
    The human ear can distinguish an echo if the reflected sound returns at least 0.10 s later. What is the minimum distance from a wall needed to hear a distinct echo? (Use 340 m/s.)
  5. Intensity from Power
    A source emits sound uniformly in all directions with power 8.0 W. What is the intensity at a distance of 2.0 m from the source?
  6. Decibel Level
    A sound has an intensity of 1.0 × 10-6 W/m2. What is its intensity level in decibels? (Take I0 = 1.0 × 10-12 W/m2.)
  7. Comparing Intensities (dB difference)
    Sound A is 30 dB and Sound B is 50 dB. How many times more intense is Sound B than Sound A?
  8. Beat Frequency
    Two tuning forks produce frequencies of 256 Hz and 262 Hz. What beat frequency is heard?
  9. Doppler Effect (Moving Source)
    A siren emits a frequency of 900 Hz. The source moves toward a stationary observer at 20 m/s. What frequency does the observer hear? (Use v = 340 m/s.)
  10. Doppler Effect (Moving Observer)
    A stationary source emits 500 Hz. An observer moves toward the source at 15 m/s. What frequency is observed? (Use v = 340 m/s.)
  11. Open Pipe Fundamental
    An open pipe has length 0.85 m. What is its fundamental frequency? (Use v = 340 m/s.)
  12. Closed Pipe Fundamental
    A pipe closed at one end has length 0.25 m. What is its fundamental frequency? (Use v = 340 m/s.)
  13. Harmonics in a Closed Pipe
    A closed pipe has a fundamental frequency of 200 Hz. What is the next allowed harmonic frequency?
  14. Standing Wave on a String
    A string fixed at both ends has length 1.20 m and vibrates in its second harmonic. What is the wavelength of the wave on the string?
  15. Frequency from Intensity Change
    A sound’s intensity increases by a factor of 100. By how many decibels does the sound level increase?

Answer Key (With Solutions)

  1. Answer: 340 m/s
    Solution: Use v = fλ = (680 Hz)(0.50 m) = 340 m/s.
  2. Answer: 2.0 m
    Solution: λ = v/f = 340/170 = 2.0 m.
  3. Answer: 0.50 s
    Solution: Echo travels to wall and back: distance = 2(85) = 170 m.
    t = d/v = 170/340 = 0.50 s.
  4. Answer: 17 m
    Solution: Need round-trip time ≥ 0.10 s.
    Round-trip distance = vt = (340)(0.10) = 34 m.
    One-way distance = 34/2 = 17 m.
  5. Answer: 0.159 W/m2
    Solution: For spherical spreading, I = P/(4πr2).
    I = 8.0 / (4π(2.0)2) = 8.0 / (16π) = 0.159 W/m2 (approx).
  6. Answer: 60 dB
    Solution: β = 10 log(I/I0) = 10 log(10-6/10-12)
    = 10 log(106) = 10(6) = 60 dB.
  7. Answer: 100 times more intense
    Solution: Δβ = 50 − 30 = 20 dB.
    Intensity ratio = 10(Δβ/10) = 102 = 100.
  8. Answer: 6 Hz
    Solution: fbeats = |f1 − f2| = |256 − 262| = 6 Hz.
  9. Answer: 956 Hz (approx)
    Solution: Moving source toward observer:
    f’ = f(v/(v − vs)) = 900(340/(340 − 20)) = 900(340/320) = 900(1.0625) = 956.25 Hz.
  10. Answer: 522 Hz (approx)
    Solution: Moving observer toward source:
    f’ = f((v + vo)/v) = 500((340 + 15)/340) = 500(355/340) = 500(1.0441) = 522.1 Hz.
  11. Answer: 200 Hz
    Solution: For open pipe fundamental: f1 = v/(2L).
    f1 = 340 / (2 × 0.85) = 340/1.70 = 200 Hz.
  12. Answer: 340 Hz
    Solution: For closed pipe fundamental: f1 = v/(4L).
    f1 = 340 / (4 × 0.25) = 340/1.0 = 340 Hz.
  13. Answer: 600 Hz
    Solution: Closed pipe supports odd harmonics: 1st, 3rd, 5th…
    Next after 200 Hz (1st) is 3rd harmonic: f3 = 3f1 = 3(200) = 600 Hz.
  14. Answer: 1.20 m
    Solution: For a string fixed at both ends: λn = 2L/n.
    For n = 2: λ = 2L/2 = L = 1.20 m.
  15. Answer: 20 dB
    Solution: Δβ = 10 log(I2/I1) = 10 log(100) = 10(2) = 20 dB.

NMAT Physics Review: NMAT Study Guide

NMAT Study Guide: Complete Preparation Guide for Medical School in the Philippines