Physics Chapter Waves & Sound

WAVES AND SOUND

FSc Pre-Medical  |  Pakistan Armed Forces Nursing Service

Chapter Outline

1. Wave Motion — Types and Properties
2. Wave Equation and Speed
3. Superposition and Interference
4. Stationary (Standing) Waves
5. Doppler Effect
6. Sound Waves — Properties and Speed
7. Intensity and Loudness
8. Practice MCQs (75 Questions)
9. Short Practice Test
10. Wave Motion — Types and Properties

A wave is a disturbance that transfers energy from one place to another without transferring matter. The medium particles oscillate about their mean positions.

1.1 Types of Waves

Transverse Waves

KEY CONCEPT:  In transverse waves, the direction of oscillation of particles is PERPENDICULAR to the direction of wave propagation. Example: light waves, water surface waves, waves on a string.

• Particles vibrate up and down while wave moves horizontally
• Can be polarised — only transverse waves show polarisation
• Can travel through solids and on surfaces of liquids

Longitudinal Waves

KEY CONCEPT:  In longitudinal waves, the direction of oscillation of particles is PARALLEL to the direction of wave propagation. Example: sound waves, compression waves in a spring.

• Particles vibrate back and forth (compression and rarefaction)
• Cannot be polarised
• Can travel through solids, liquids and gases

1.2 Key Wave Properties

QUICK TIP:  T = 1/f (period = 1/frequency). High frequency = short period = short wavelength (for same speed).

2. Wave Equation and Speed

KEY CONCEPT:  Wave speed v = fλ (frequency × wavelength). Also v = λ/T. The most important wave equation.

2.1 The Wave Equation

• v = fλ — the universal wave equation
• v = λ/T (since f = 1/T)
• For a given medium, wave speed is constant
• If frequency increases, wavelength decreases (speed stays same)

EXAMPLE:  A wave has frequency 500 Hz and wavelength 0.68 m. Speed = fλ = 500 × 0.68 = 340 m/s. (Speed of sound in air!)

QUICK TIP:  v = fλ. If two quantities given, find third. Example: f=256 Hz, v=340 m/s → λ = v/f = 340/256 = 1.33 m.

2.2 Speed of Sound in Different Media

KEY CONCEPT:  Speed of sound: Steel > Water > Air. Sound cannot travel in vacuum. Speed increases with temperature: v increases by 0.61 m/s per 1°C.

3. Superposition and Interference

KEY CONCEPT:  Superposition: When two or more waves meet, the resultant displacement is the algebraic sum of their individual displacements.

3.1 Constructive Interference

KEY CONCEPT:  Constructive interference: waves meet IN PHASE (path difference = nλ). Amplitudes ADD UP — maximum amplitude.

• Condition: path difference = 0, λ, 2λ, 3λ…

EXAMPLE:  Path difference = 2λ → constructive interference → loudest sound.

3.2 Destructive Interference

KEY CONCEPT:  Destructive interference: waves meet OUT OF PHASE (path difference = (n+½)λ). Amplitudes CANCEL — zero amplitude.

• Condition: path difference = λ/2, 3λ/2, 5λ/2…

EXAMPLE:  Path difference = λ/2 → destructive interference → silence.

QUICK TIP:  Constructive = crest meets crest (same phase). Destructive = crest meets trough (opposite phase).

3.3 Beats

KEY CONCEPT:  Beats are produced by superposition of two waves with slightly different frequencies. Beat frequency = |f₁ − f₂|.

EXAMPLE:  Two tuning forks: 440 Hz and 436 Hz. Beat frequency = |440 − 436| = 4 Hz. You hear 4 loud-soft cycles per second.

4. Stationary (Standing) Waves

Standing waves form when two identical waves travel in opposite directions and superpose. The wave pattern does not appear to move.

4.1 Nodes and Antinodes

KEY CONCEPT:  Node: point of zero displacement (destructive interference permanently). Antinode: point of maximum displacement. Distance between adjacent nodes = λ/2.

• Nodes: particles never move
• Antinodes: particles have maximum amplitude
• Distance between consecutive node and antinode = λ/4
• Distance between two consecutive nodes = λ/2

4.2 Harmonics in a String (fixed at both ends)

• Fundamental (1st harmonic): f₁ = v/2L
• 2nd harmonic: f₂ = 2f₁ = v/L
• nth harmonic: fn = nf₁
• All harmonics present (1, 2, 3, 4…)

EXAMPLE:  String: L=1m, v=200 m/s. f₁ = 200/(2×1) = 100 Hz. f₂ = 200 Hz, f₃ = 300 Hz.

4.3 Harmonics in Pipes

Open Pipe (open both ends)

• f₁ = v/2L — All harmonics present

Closed Pipe (closed one end)

• f₁ = v/4L — Only ODD harmonics (1st, 3rd, 5th…)

KEY CONCEPT:  Open pipe: all harmonics. Closed pipe: odd harmonics only. Same length: closed pipe fundamental = half of open pipe.

EXAMPLE:  Closed pipe L=0.5m, v=340m/s: f₁ = 340/(4×0.5) = 170 Hz. Open same length: f₁ = 340/1.0 = 340 Hz.

Chapter 15 MCQs 

Q50. Doppler effect application:

• (A) rainbow
• (B) medical ultrasound of blood flow
• (C) shadow
• (D) refraction

Answer: (B) medical ultrasound of blood flow

Explanation: Doppler ultrasound detects frequency shift to measure blood flow velocity.

Q51. Human audible frequency range:

• (A) 0-20 Hz
• (B) 20-20,000 Hz
• (C) 20-200,000 Hz
• (D) 200-2000 Hz

Answer: (B) 20-20,000 Hz

Explanation: Human hearing: 20 Hz to 20 kHz.

Q52. Ultrasound frequency:

• (A) below 20 Hz
• (B) 20-20,000 Hz
• (C) above 20,000 Hz
• (D) exactly 20,000 Hz

Answer: (C) above 20,000 Hz

Explanation: Ultrasound = above 20 kHz = above human hearing.

Q53. Infrasound frequency:

• (A) below 20 Hz
• (B) 20-20,000 Hz
• (C) above 20,000 Hz
• (D) 20 Hz exactly

Answer: (A) below 20 Hz

Explanation: Infrasound = below 20 Hz. Produced by earthquakes, large animals.

Q54. Pitch of sound depends on:

• (A) amplitude
• (B) frequency
• (C) speed
• (D) wavelength

Answer: (B) frequency

Explanation: Pitch (how high or low) depends on frequency. High frequency = high pitch.

Q55. Loudness depends on:

• (A) frequency
• (B) wavelength
• (C) amplitude/intensity
• (D) speed

Answer: (C) amplitude/intensity

Explanation: Loudness depends on amplitude/intensity. Larger amplitude = louder.

Q56. Minimum distance for echo (v=340, t=0.1s):

• (A) 34 m
• (B) 17 m
• (C) 8.5 m
• (D) 68 m

Answer: (B) 17 m

Explanation: d = v×t/2 = 340×0.1/2 = 17 m.

Q57. SONAR uses:

• (A) visible light
• (B) infrared
• (C) ultrasound reflection
• (D) radio waves

Answer: (C) ultrasound reflection

Explanation: SONAR sends ultrasound pulses and measures echo time to find depth.

Q58. Intensity I ∝ amplitude to power:

• (A) 1
• (B) 2
• (C) 3
• (D) 0.5

Answer: (B) 2

Explanation: I ∝ A². Double amplitude → intensity ×4.

Q59. SI unit of intensity:

• (A) dB
• (B) Hz
• (C) W/m²
• (D) N/m

Answer: (C) W/m²

Explanation: Intensity = Power/Area = W/m².

Q60. If distance doubles from sound source, intensity becomes:

• (A) double
• (B) half
• (C) 1/4
• (D) same

Answer: (C) 1/4

Explanation: I ∝ 1/r². Double r → I = I/4.

Q61. Threshold of hearing I₀ =

• (A) 10⁻⁶ W/m²
• (B) 10⁻¹² W/m²
• (C) 10⁻³ W/m²
• (D) 1 W/m²

Answer: (B) 10⁻¹² W/m²

Explanation: I₀ = 10⁻¹² W/m² = 0 dB reference level.

Q62. 10 dB increase means intensity increases by:

• (A) 2×
• (B) 5×
• (C) 10×
• (D) 100×

Answer: (C) 10×

Explanation: Every 10 dB = 10× more intensity (logarithmic scale).

Q63. Normal conversation dB level:

• (A) 20 dB
• (B) 40 dB
• (C) 60 dB
• (D) 80 dB

Answer: (C) 60 dB

Explanation: Normal conversation ≈ 60 dB. Whisper ≈ 30 dB.

Q64. Redshift in astronomy means star is:

• (A) approaching
• (B) stationary
• (C) moving away
• (D) exploding

Answer: (C) moving away

Explanation: Redshift = lower frequency = longer wavelength = source moving away.

Q65. Blueshift means observed frequency is:

• (A) lower
• (B) same
• (C) higher
• (D) zero

Answer: (C) higher

Explanation: Blueshift = higher frequency = source moving toward observer.

Q66. Reverberation is:

• (A) single echo
• (B) multiple reflections in enclosed space
• (C) Doppler shift
• (D) resonance

Answer: (B) multiple reflections in enclosed space

Explanation: Reverberation = persistence of sound due to multiple reflections in closed spaces.

Q67. Quality (timbre) of sound depends on:

• (A) frequency only
• (B) amplitude only
• (C) harmonics/waveform
• (D) speed

Answer: (C) harmonics/waveform

Explanation: Timbre = quality distinguishing same note from different instruments. Due to harmonic content.

Q68. Bat echolocation uses:

• (A) infrasound
• (B) audible sound
• (C) ultrasound
• (D) radio waves

Answer: (C) ultrasound

Explanation: Bats emit ultrasound (>20 kHz) and use echoes for navigation and hunting.

Q69. Ship sonar echo in 4s (v=1500 m/s). Sea depth:

• (A) 6000 m
• (B) 3000 m
• (C) 1500 m
• (D) 750 m

Answer: (B) 3000 m

Explanation: Depth = v×t/2 = 1500×4/2 = 3000 m.

Q70. 20 dB increase from 50 dB gives:

• (A) 60 dB
• (B) 70 dB
• (C) 100 dB
• (D) 1000 dB

Answer: (B) 70 dB

Explanation: 50 + 20 = 70 dB. (Intensity increases 100×.)

Q71. Speed of sound in water vs air: water is ___ air:

• (A) same as
• (B) slower than
• (C) faster than
• (D) depends on frequency

Answer: (C) faster than

Explanation: Water (~1500 m/s) is about 4× faster than air (~340 m/s).

Q72. When sound goes from air to steel, frequency:

• (A) increases
• (B) decreases
• (C) stays same
• (D) doubles

Answer: (C) stays same

Explanation: At boundaries, frequency is conserved. Speed and wavelength change.

Q73. Ambulance speed 20 m/s, siren 600 Hz, v=340 m/s. Moving away. Observed f’:

• (A) 565 Hz
• (B) 600 Hz
• (C) 636 Hz
• (D) 800 Hz

Answer: (A) 565 Hz

Explanation: f’ = fv/(v+vs) = 600×340/(340+20) = 204000/360 = 566.7 ≈ 565 Hz.

Q74. Which has LOWEST frequency in EM spectrum?

• (A) gamma rays
• (B) X-rays
• (C) visible light
• (D) radio waves

Answer: (D) radio waves

Explanation: Radio waves have lowest frequency (longest wavelength) in the EM spectrum.

Q75. Intensity at 6 m from a speaker if intensity at 3 m is 100 W/m²:

• (A) 25 W/m²
• (B) 50 W/m²
• (C) 100 W/m²
• (D) 200 W/m²

Answer: (A) 25 W/m²

Explanation: I ∝ 1/r². Distance doubles (3→6) → I becomes 100/4 = 25 W/m².