Chapter Summary
What is Oscillatory Motion?
Oscillation is a periodic motion where an object moves back and forth at regular intervals about its equilibrium position. Examples include pendulum motion, swing motion, and vibrating tuning forks.
Key Definitions and Concepts
1. Amplitude (a)
- The maximum displacement from equilibrium position
- SI unit: metre (m)
- Symbol: a
2. Period (T)
- Time taken for one complete oscillation
- SI unit: second (s)
- Symbol: T
3. Frequency (f)
- Number of oscillations per second
- SI unit: hertz (Hz)
- Symbol: f
- Relationship: f = 1/T
4. Natural Frequency
- The frequency at which an object vibrates freely
- Depends on: length, size, elasticity, and nature of material
Wave Motion
Wave motion is the transfer of energy from one part of a medium to another through oscillations, without the actual displacement of particles.
Types of Waves
Mechanical Waves
- Require a medium for transmission
- Two types: Longitudinal and Transverse
Electromagnetic Waves
- Do not require a medium
- Examples: radio waves, light waves, X-rays
- All electromagnetic waves are transverse
Wave Classifications
Longitudinal Waves
- Particles vibrate parallel to wave propagation direction
- Form compressions (high pressure) and rarefactions (low pressure)
- Examples: Sound waves, seismic P-waves
- Sound is a longitudinal wave
Transverse Waves
- Particles vibrate perpendicular to wave propagation direction
- Form crests (highest points) and troughs (lowest points)
- Examples: Light waves, electromagnetic waves
- No pressure variations occur
Wave Characteristics
Wavelength (λ)
- Distance between two consecutive particles in same phase
- Distance between consecutive crests/troughs (transverse)
- Distance between consecutive compressions/rarefactions (longitudinal)
- SI unit: metre (m)
Wave Speed (v)
- Distance traveled by wave per second
- Formula: v = fλ (speed = frequency × wavelength)
- SI unit: m/s
Sound Wave Properties
Speed of Sound
- In air at room temperature: ~350 m/s
- In water: ~1480 m/s
- Faster in denser media
Audible Range for Humans
- Audible sound: 20 Hz to 20,000 Hz (20 kHz)
- Infrasonic: Below 20 Hz
- Ultrasonic: Above 20,000 Hz
Sound Phenomena
Reflection of Sound
- Sound waves bounce back when hitting surfaces
- Smooth surfaces reflect better than rough surfaces
- Applications: soundboards, curved hall ceilings
Echo
- Reflected sound heard distinctly after original sound
- Minimum distance for echo: 17.5 m (in air)
- Formula for distance: d = vt/2
Reverberation
- Lingering of sound due to multiple reflections
- Gradually fades away
- Controlled in halls using rough surfaces
Forced Vibration
- Vibration induced by external vibrating object
- Example: table vibrating when mixie operates on it
Resonance
- Maximum amplitude vibration when natural frequencies match
- Applications: musical instruments, MRI, radio tuning
Ultrasonic Applications
Medical Uses
- Kidney stone crushing
- Physiotherapy
- Ultrasonography (imaging internal organs)
Industrial Uses
- Cleaning electronic components
- SONAR (underwater distance measurement)
Biological Uses
- Bat echolocation
- Animal communication
Important Formulas
- Frequency-Period Relationship: f = 1/T
- Wave Speed: v = fλ
- Echo Distance: d = vt/2
- Minimum Echo Distance: d = v/(2 × 10) = 17.5 m (in air)
Question and Answer Study Material
Section A: Oscillatory Motion and Basic Concepts
Q1. What is oscillatory motion? Give examples.
Answer: Oscillatory motion is a periodic motion in which an object moves to and fro at regular intervals of time about its equilibrium position.
Examples:
- Motion of a pendulum in a clock
- Swing motion
- Vibrating tuning fork
- Vibrating guitar strings
Q2. Define amplitude, period, and frequency. Write their SI units.
Answer:
- Amplitude (a): The magnitude of maximum displacement to one side from equilibrium position. SI unit: metre (m)
- Period (T): The time taken for one complete oscillation. SI unit: second (s)
- Frequency (f): The number of oscillations completed in one second. SI unit: hertz (Hz)
Q3. What is the relationship between period and frequency?
Answer: Period and frequency are inversely related: f = 1/T
As period increases, frequency decreases and vice versa.
Q4. A pendulum takes 1 minute to complete 30 oscillations. Calculate its period and frequency.
Answer: Given: 30 oscillations in 60 seconds
Period (T) = Total time/Number of oscillations = 60s/30 = 2s
Frequency (f) = 1/T = 1/2 = 0.5 Hz
Q5. What is natural frequency? What factors influence it?
Answer: Natural frequency is the frequency at which an object vibrates freely when disturbed.
Factors influencing natural frequency:
- Length of the object
- Size of the object
- Elasticity of the material
- Nature of the material
Section B: Wave Motion and Types
Q6. What is wave motion? How does energy transfer occur in waves?
Answer: Wave motion is the continuous propagation of energy from one part of a medium to other parts through oscillations, without actual displacement of the medium particles.
Energy transfer occurs when disturbance in one part spreads to adjacent parts, which then transfer energy to their neighboring parts, and so on.
Q7. Distinguish between mechanical and electromagnetic waves.
Answer:
Mechanical Waves:
- Require medium for transmission
- Examples: sound, seismic waves
- Can be longitudinal or transverse
- Speed depends on medium properties
Electromagnetic Waves:
- Do not require medium
- Examples: light, radio waves, X-rays
- Always transverse
- Travel at speed of light in vacuum
Q8. Compare longitudinal and transverse waves.
Answer:
Longitudinal Waves:
- Particles vibrate parallel to wave direction
- Form compressions and rarefactions
- Pressure variations occur
- Example: Sound waves
Transverse Waves:
- Particles vibrate perpendicular to wave direction
- Form crests and troughs
- No pressure variations
- Example: Light waves
Q9. Why is sound called a longitudinal wave?
Answer: Sound is called a longitudinal wave because:
- Air particles vibrate parallel to the direction of sound propagation
- Sound creates alternating regions of high pressure (compressions) and low pressure (rarefactions)
- These pressure variations travel through the medium as the sound wave propagates
Section C: Wave Characteristics
Q10. Define wavelength. How is it measured in longitudinal and transverse waves?
Answer: Wavelength (λ) is the distance between two consecutive particles in the same phase of vibration.
Measurement:
- Transverse waves: Distance between consecutive crests or troughs
- Longitudinal waves: Distance between consecutive compressions or rarefactions
Q11. Derive the relationship between wave speed, frequency, and wavelength.
Answer: When a wave travels one wavelength (λ) distance, it takes one period (T) time.
Speed = Distance/Time v = λ/T
Since f = 1/T: v = fλ
This is the fundamental wave equation.
Q12. A wave travels 700 m in 2 seconds. If its frequency is 50 Hz, find its wavelength.
Answer: Given: Distance = 700 m, Time = 2 s, Frequency = 50 Hz
Speed (v) = Distance/Time = 700/2 = 350 m/s
Using v = fλ: 350 = 50 × λ λ = 350/50 = 7 m
Q13. What is the relationship between wavelength and frequency when wave speed is constant?
Answer: When wave speed is constant: v = fλ = constant
Therefore: f ∝ 1/λ
Frequency is inversely proportional to wavelength. As frequency increases, wavelength decreases and vice versa.
Section D: Sound Properties
Q14. What is the speed of sound in different media? Why does it vary?
Answer: Speed of sound:
- Air (room temperature): ~350 m/s
- Water: ~1480 m/s
- Steel: ~5000 m/s
Sound speed varies because it depends on:
- Density of the medium
- Elasticity of the medium
- Temperature
Generally, sound travels faster in denser and more elastic media.
Q15. Define the audible range for humans. What are infrasonic and ultrasonic sounds?
Answer: Audible range for humans: 20 Hz to 20,000 Hz (20 kHz)
Infrasonic sounds:
- Frequency below 20 Hz
- Examples: Earthquake waves, elephant communication
Ultrasonic sounds:
- Frequency above 20,000 Hz
- Examples: Bat echolocation, medical ultrasound
Q16. A sound wave has frequency 175 Hz and wavelength 2 m. Calculate the speed of sound.
Answer: Given: f = 175 Hz, λ = 2 m
Using v = fλ: v = 175 × 2 = 350 m/s
Section E: Sound Phenomena
Q17. What is reflection of sound? How do smooth and rough surfaces affect it?
Answer: Reflection of sound occurs when sound waves bounce back after striking a surface.
Smooth surfaces: Reflect sound more effectively, producing clearer reflections Rough surfaces: Scatter sound waves, reducing the intensity of reflection
Applications: Soundboards, curved hall ceilings use smooth surfaces for better sound distribution.
Q18. What is echo? What conditions are necessary to hear an echo?
Answer: Echo is the reflected sound heard distinctly after the original sound.
Conditions for echo:
- Reflecting surface must be at least 17.5 m away
- Time gap between original and reflected sound should be at least 0.1 seconds
- Persistence of hearing limits our ability to distinguish sounds
Q19. Calculate the minimum distance for echo formation. (Speed of sound = 350 m/s)
Answer: For distinct echo, minimum time gap = 0.1 s (persistence of hearing)
In this time, sound travels to reflecting surface and back: Total distance = Speed × Time = 350 × 0.1 = 35 m
Minimum distance to reflecting surface = 35/2 = 17.5 m
Q20. Differentiate between echo and reverberation.
Answer:
Echo:
- Distinct reflection of sound
- Heard after original sound ceases
- Requires minimum 17.5 m distance
- Single reflection
- Example: Mountain echo
Reverberation:
- Multiple reflections causing sound to linger
- Overlaps with original sound
- Occurs in enclosed spaces
- Multiple reflections
- Example: Sound in large halls
Section F: Forced Vibration and Resonance
Q21. What is forced vibration? Give examples.
Answer: Forced vibration is the vibration of an object induced by an external vibrating object.
Examples:
- Table vibrating when a mixie operates on it
- Tuning fork making a table vibrate when pressed against it
- Building vibrations during earthquakes
Q22. Explain resonance with examples.
Answer: Resonance occurs when the natural frequency of a forced object equals the natural frequency of the forcing object, resulting in maximum amplitude vibrations.
Examples:
- Hacksaw blades of same length vibrating together
- Radio tuning to specific frequencies
- Musical instruments producing specific notes
- MRI scanning technology
Q23. What are the applications of forced vibration and resonance?
Answer: Applications include:
- Medical: MRI scanning, stethoscope operation
- Communication: Radio tuning, television transmission
- Musical instruments: Guitar, violin, veena, harmonium
- Sound amplification: Megaphones, horns, trumpets
Section G: Ultrasonic Applications
Q24. List the medical applications of ultrasonic waves.
Answer: Medical applications:
- Kidney stone treatment: Crushing small stones using focused ultrasound
- Physiotherapy: Treatment of muscle and joint problems
- Ultrasonography: Creating images of internal organs (kidney, liver, gallbladder, uterus)
- Diagnosis: Detecting abnormalities in body tissues
Q25. How does ultrasonography work?
Answer: In ultrasonography:
- Ultrasonic waves are transmitted into the body
- These waves travel through tissues and reflect at areas of varying density
- Reflected waves return to the transducer
- The reflected signals are converted into electrical signals
- These signals are processed to form images of internal organs on a screen
Q26. What is SONAR? How does it work?
Answer: SONAR (Sound Navigation and Ranging) is a device that uses ultrasonic waves to find distances to underwater objects.
Working principle:
- Ultrasonic transmitter on ship sends waves downward
- Waves reflect from underwater objects (sea floor, rocks, submarines)
- Reflected waves are detected by receiver
- Distance is calculated using: d = vt/2 (where t is total travel time)
Q27. An ultrasonic wave returns after 0.2 s from a rock at the sea bottom. Find the distance if wave speed in seawater is 1522 m/s.
Answer: Given: Time = 0.2 s, Speed = 1522 m/s
Using d = vt/2: d = (1522 × 0.2)/2 = 304.4/2 = 152.2 m
Section H: Destructive Waves
Q28. What are seismic waves? How are they related to earthquakes?
Answer: Seismic waves are waves that travel through Earth's crust due to:
- Earthquakes
- Volcanic eruptions
- Massive explosions
Relationship to earthquakes:
- Earthquakes generate seismic waves
- These waves cause ground shaking and structural damage
- Earthquake intensity is measured using the Richter scale
- Seismology is the study of these waves
Q29. What is tsunami? How is it related to seismic activity?
Answer: Tsunami is a series of gigantic ocean waves caused by displacement of large volumes of seawater.
Relationship to seismic activity:
- Underwater earthquakes can trigger tsunamis
- Coastal earthquakes may also cause tsunamis
- Sudden displacement of sea floor during earthquakes creates massive water movements
- These movements propagate as tsunami waves across oceans
Q30. What safety measures should be taken during tsunamis?
Answer: Safety measures:
- Follow official tsunami warning center instructions
- Move to higher ground immediately when warning is issued
- Stay away from beaches and coastal areas
- Do not return to danger zones until all-clear is given
- Have emergency evacuation plans ready
- Stay informed through official channels
Section I: Numerical Problems
Q31. A longitudinal wave traveling at 350 m/s has frequency 35 Hz. Find: a) Distance between consecutive compressions b) Distance between consecutive rarefactions
Answer: Given: v = 350 m/s, f = 35 Hz
Using v = fλ: λ = v/f = 350/35 = 10 m
a) Distance between consecutive compressions = λ = 10 m b) Distance between consecutive rarefactions = λ = 10 m
Q32. A firecracker's echo is heard after 1 s. Find the distance to the reflecting surface. (v = 350 m/s)
Answer: Given: Time = 1 s, Speed = 350 m/s
Using d = vt/2: d = (350 × 1)/2 = 175 m
The reflecting surface is 175 m away.
Q33. What is the minimum distance for echo in water? (Speed of sound in water = 1480 m/s)
Answer: For echo, minimum time gap = 0.1 s
Total distance sound travels = 1480 × 0.1 = 148 m Minimum distance to reflecting surface = 148/2 = 74 m
Q34. A bat can hear frequencies up to 120 kHz. Find the minimum wavelength it can detect. (v = 350 m/s)
Answer: Given: f = 120 kHz = 120,000 Hz, v = 350 m/s
Using v = fλ: λ = v/f = 350/120,000 = 0.00292 m = 2.92 mm
Q35. If a pendulum has frequency 1 Hz, what is its period? If another pendulum takes 0.5 s for one oscillation, what is its frequency?
Answer: For first pendulum: Given: f = 1 Hz T = 1/f = 1/1 = 1 s
For second pendulum: Given: T = 0.5 s
f = 1/T = 1/0.5 = 2 Hz
Quick Revision Points
Key Formulas to Remember
- f = 1/T
- v = fλ
- d = vt/2 (for echo)
- Minimum echo distance = 17.5 m (in air)
Important Values
- Speed of sound in air: 350 m/s
- Speed of sound in water: 1480 m/s
- Human audible range: 20 Hz to 20 kHz
- Persistence of hearing: 0.1 s
Wave Types
- Longitudinal: Sound, compressions & rarefactions, parallel vibration
- Transverse: Light, crests & troughs, perpendicular vibration
Applications to Remember
- Ultrasound: Medical imaging, kidney stone treatment, SONAR
- Resonance: Musical instruments, radio tuning, MRI
- Echo: Distance measurement, communication