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Unit notes and practice questions

Grade 9 Physics unit 6: Mechanical Oscillation and Sound

Everything here is for unit 6 of the Grade 9 Physics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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10
Cards
10
Practice questions on unit 6
114–139
The textbook pages for unit 6

What unit 6 covers

  1. 1.Common Characteristics of Waves
  2. 2.String, Pendulum and Spring
  3. 3.Propagation of Waves and Energy Transmission
  4. 4.Sound Waves
  5. 5.Superposition of Waves
  6. 6.Characteristics of Sound Waves

Short notes for unit 6

The cards filed under this chapter on the Grade 9 Physics page. Open one at a time.

When you use it

Apply when solving problems where sound travels to an obstacle and reflects back to the observer or source.

Watch out

The measured time t is for the round trip. The one way distance to the obstacle is d = v times t / 2. Forgetting to divide by 2 gives double the actual distance.

Drafted from Grade 9 Physics, pages 114-139, then checked twice before it went up

T=2πmkT = 2\pi \sqrt{\frac{m}{k}}
TT
Periodss
mm
Masskgkg
kk
Spring constantN m−1N\,m^{-1}

When you use it

Use this formula to calculate the period of oscillation for a mass attached to a horizontal or vertical spring.

Watch out

Mass must be in kilograms and spring constant k in N/m. The period depends only on mass and spring constant, not on displacement amplitude.

Drafted from Grade 9 Physics, pages 114-139, then checked twice before it went up

T=2πLgT = 2\pi \sqrt{\frac{L}{g}}
TT
Periodss
LL
Length of pendulummm
gg
Acceleration due to gravitym s−2m\,s^{-2}

When you use it

Use this formula to find the period of a simple pendulum or to calculate local gravitational acceleration from pendulum length and period.

Watch out

The mass of the bob and the amplitude do not affect the period. Length must be converted from centimeters to meters.

Drafted from Grade 9 Physics, pages 114-139, then checked twice before it went up

v=3311+Tc273v = 331 \sqrt{1 + \frac{T_c}{273}}
vv
Speed of soundm s−1m\,s^{-1}
TcT_c
Air temperature∘C^\circ\text{C}

When you use it

Use this formula to find the speed of sound in air at any given Celsius temperature.

Watch out

Insert the temperature directly in degrees Celsius, not Kelvin. The speed of sound at 0 degrees Celsius is 331 m/s.

Drafted from Grade 9 Physics, pages 114-139, then checked twice before it went up

v=fλv = f \lambda
vv
Wave speedm s−1m\,s^{-1}
ff
FrequencyHzHz
λ\lambda
Wavelengthmm

When you use it

Use this formula to relate the propagation speed of any periodic wave to its frequency and wavelength.

Watch out

Wavelength must be in meters. If given the period T instead of frequency, use f = 1/T.

Drafted from Grade 9 Physics, pages 114-139, then checked twice before it went up

What to compareTransverse waveLongitudinal wave
How the particles moveAt right angles to the direction the wave travels.Back and forth along the same line the wave travels.
What the diagram showsA sine curve of crests and troughs, with the travel arrow drawn along the horizontal axis.A band of vertical lines that are alternately crowded and spread out, labelled C R C R C, with the travel arrow drawn along the band.
The parts you labelCrests at the top of the curve and troughs at the bottom.Compressions, marked C, where the particles are crowded and the pressure is high; rarefactions, marked R, where they are spread apart and the pressure is low.
ExamplesWater waves, where the surface moves up and down while the wave travels across it. Light.Sound in air, sent out as a vibrating object pushes and pulls the air beside it.
What actually travelsThe energy goes forward. Each particle only swings about its own resting place.The energy goes forward. Each particle only swings about its own resting place.

When you use it

Use when a question asks which kind a given wave is, or asks you to label a wave diagram. Compare two directions: the way the particles of the medium vibrate and the way the wave itself travels. At right angles means transverse, along the same line means longitudinal.

Watch out

Students memorise the picture instead of the motion, so they see sound drawn as a smooth curve on an oscilloscope and answer that sound is transverse. That curve is a graph of displacement or pressure against distance or time. It is not a picture of what the air is doing. In air the particles move back and forth along the very line the sound is travelling, so sound stays longitudinal however it is graphed. Judge by the direction the particles vibrate and never by the shape of the drawing.

What to compareMechanical waveElectromagnetic wave
A medium to travel throughNeeded. It must have a solid, a liquid or a gas.Not needed. It carries itself across empty space, which is how starlight reaches us, and it can still pass through glass or water.
What moves it alongEach particle of the medium pushes and pulls the next one.An oscillating electric field and magnetic field that keep regenerating each other.
SpeedThe medium sets it. Sound in air at room temperature is about 340 m/s.Every one of them travels at about 300 000 000 m/s in a vacuum.
Transverse or longitudinalEither. A wave on a string is transverse and sound in air is longitudinal.Always transverse.
ExamplesSound, water waves, a wave on a stretched string, seismic waves.Light, radio waves, X-rays. Radio, television and mobile signals all travel this way.
In medical imagingUltrasound scanning, which is sound above 20 kHz.X-ray imaging.
The test that settles itRing an alarm clock inside a jar and pump the air out. The sound fades to nothing.The light from that same clock still reaches your eye.

When you use it

Use when a question asks which waves can cross a vacuum, or gives you a list of waves to sort into two groups. One thing decides every case: a mechanical wave needs matter to travel through and an electromagnetic wave does not.

Watch out

Students see X-ray and ultrasound printed together under medical imaging and file both as electromagnetic. Ultrasound is sound. It is a mechanical wave that needs tissue to travel through, and that is exactly why the technician squeezes gel between the probe and the skin: a thin layer of air would reflect the sound back and the scan would show nothing. X-rays need no medium at all. The exam question that catches this is always the same one, asking which of the listed waves can travel through a vacuum. Light, radio and X-rays can. Sound, ultrasound, water waves and seismic waves cannot.

node to node=λ2,  node to antinode=λ4\text{node to node} = \dfrac{\lambda}{2}, \; \text{node to antinode} = \dfrac{\lambda}{4}
λ\lambda
the wavelength of each of the two travelling waves that build the patternm\mathrm{m}
AA
the amplitude of one of those two waves on its ownm\mathrm{m}
2A2A
the amplitude at an antinode, twice that of a single wavem\mathrm{m}

When you use it

Use for a string, a rope or an air column vibrating in a fixed pattern, where a node stays still at all times and an antinode swings twice as far as either wave alone. Node to node is half a wavelength and node to antinode is a quarter, which is what turns a measured length into a wavelength.

Watch out

Students take the distance between two neighbouring nodes to be one full wavelength. It is half of one. That single slip breaks every string and pipe question: a string of length L vibrating in its fundamental holds one loop, so L is half a wavelength, the wavelength is twice L, and the frequency comes out as the wave speed divided by twice L. Read the wavelength as L instead and the answer is double the true frequency every time. Count loops rather than nodes, because each loop between two nodes is half a wave. One more thing to be exact about: an antinode has the largest amplitude of swing, and its displacement still passes through zero twice every cycle, at the instants when the whole medium is momentarily straight.

Ares=A1+A2 or Ares=∣A1−A2∣A_{res} = A_{1} + A_{2} \text{ or } A_{res} = |A_{1} - A_{2}|
A1A_{1}
the amplitude of the first wavem\mathrm{m}
A2A_{2}
the amplitude of the second wavem\mathrm{m}
AresA_{res}
the resultant amplitude where the two overlap: the sum when they arrive in phase, the difference when they arrive half a cycle apartm\mathrm{m}

When you use it

Use when two waves overlap at a point and a question asks how loud, how bright or how tall the result is. Add the amplitudes when the waves arrive in phase, which means a path difference of a whole number of wavelengths, and subtract them when the waves arrive half a cycle apart, which means a path difference of an odd number of half wavelengths.

Watch out

Students read destructive as the waves being destroyed and expect silence or darkness whenever two waves meet out of step. Two things have to be true before the result is zero: the waves must arrive exactly half a cycle apart, which is an odd number of half wavelengths of path difference rather than merely out of step, and their amplitudes must be equal. A 3 cm wave meeting a 2 cm wave in antiphase leaves a 1 cm wave. Nothing is lost either. The energy taken from the quiet points reappears at the loud ones, and both waves come out of the meeting exactly as they went in.

E∝A2f2E \propto A^{2} f^{2}
EE
the energy the wave carries. It travels forward while the particles of the medium only swing about their resting placesJ\mathrm{J}
AA
the displacement amplitude: how far each particle swings from its resting placem\mathrm{m}
ff
the frequency of the wave, the number of complete oscillations each secondHz\mathrm{Hz}

When you use it

Use when a question changes the amplitude or the frequency of a sound, a wave on a string or a water wave and asks what happens to the energy it carries. Both quantities are squared, so doubling either one on its own gives four times the energy and doubling both gives sixteen times.

Watch out

Students read the proportion as a straight multiplication and say that doubling the amplitude doubles the energy. It quadruples, because the amplitude is squared, and doubling the amplitude and the frequency together gives sixteen times the energy rather than four. The second trap is hearing higher frequency as louder: a higher frequency sound is higher in pitch, and loudness is what the amplitude controls. Note also that the relation carries no constant, so it compares two waves of the same kind in the same medium and never gives an energy in joules. It belongs to mechanical waves alone and does not carry over to light, whose brightness is set by amplitude while frequency changes the colour.

Practice questions on unit 6

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

What is the name for the maximum displacement of an oscillating object from its equilibrium position?

  1. A.Wavelength
  2. B.Amplitude
  3. C.Frequency
  4. D.Period
Show the answer and why

Answer: B Amplitude

Why

Amplitude is defined as the maximum displacement from the equilibrium (rest) position.

Question 2

Which type of wave does not require a material medium and can travel through a vacuum?

  1. A.Sound wave
  2. B.Water wave
  3. C.Wave on a string
  4. D.Electromagnetic wave
Show the answer and why

Answer: D Electromagnetic wave

Why

Electromagnetic waves (such as light and radio waves) consist of vibrating electric and magnetic fields and do not require a material medium to travel.

Question 3

When a musician in Addis Ababa plucks a kirar string, which wave characteristic determines the pitch of the sound produced?

  1. A.Frequency
  2. B.Amplitude
  3. C.Wave speed
  4. D.Loudness
Show the answer and why

Answer: A Frequency

Why

The pitch of a sound depends directly on the frequency of the sound wave. Higher frequency produces a higher pitch.

Question 4

Through which of the following materials does sound travel the fastest?

  1. A.Air
  2. B.Water
  3. C.Iron
  4. D.Rubber
Show the answer and why

Answer: C Iron

Why

Sound travels fastest in solids because particles are close together. In iron, the speed of sound is 5950 m/s, which is much faster than in liquids or gases.

Question 5

A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. What is the speed of this wave?

  1. A.170 m/s
  2. B.340 m/s
  3. C.735 m/s
  4. D.250 m/s
Show the answer and why

Answer: B 340 m/s

Why

Wave speed is calculated using v = f × λ. Therefore, v = 500 Hz × 0.68 m = 340 m/s.

Question 6

A student in Hawassa doubles the mass of the bob of a simple pendulum while keeping its string length unchanged. What happens to the period of oscillation?

  1. A.It doubles
  2. B.It is halved
  3. C.It remains unchanged
  4. D.It increases by four times
Show the answer and why

Answer: C It remains unchanged

Why

The period of a simple pendulum depends only on its length (L) and acceleration due to gravity (g). It is completely independent of the mass of the bob.

Question 7

An applied force of 20 N stretches a spring by 0.05 m. What is the spring constant (k) of the spring?

  1. A.1 N/m
  2. B.100 N/m
  3. C.400 N/m
  4. D.200 N/m
Show the answer and why

Answer: C 400 N/m

Why

According to Hooke's law, k = Fapp / x. Therefore, k = 20 N / 0.05 m = 400 N/m.

Question 8

A person stands facing a cliff in Gondar and claps their hands. The echo is heard 2 seconds later. If the speed of sound is 343 m/s, what is the distance to the cliff?

  1. A.171.5 m
  2. B.343 m
  3. C.686 m
  4. D.514.5 m
Show the answer and why

Answer: B 343 m

Why

The sound travels to the cliff and back in 2 s, so the time taken to reach the cliff is 1 s. Distance = speed × time = 343 m/s × 1 s = 343 m.

Question 9

A block of mass 4 kg is connected to a spring and oscillates with a period of (π/10) s. What is the spring constant of the spring?

  1. A.160 N/m
  2. B.400 N/m
  3. C.800 N/m
  4. D.1600 N/m
Show the answer and why

Answer: D 1600 N/m

Why

The period formula is T = 2π√(m/k). Substituting the values: π/10 = 2π√(4/k) gives 1/20 = 2/√k, which means √k = 40, so k = 1600 N/m.

Question 10

In a standing wave on a string, what is the distance between a node and its nearest adjacent antinode?

  1. A.λ / 4
  2. B.λ / 2
  3. C.λ
  4. D.2λ
Show the answer and why

Answer: A λ / 4

Why

For a standing wave, the distance between two adjacent nodes or two adjacent antinodes is λ/2, while the distance between a node and an adjacent antinode is λ/4.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 6

Unit 6 runs from page 114 to page 139 of the Ministry of Education edition.

Ministry of Education edition, English, 179 pages

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