Physics Sound, Light, Circular Motion and Planets
Formula for fundamental frequency
( f = frequency; l = length of string; T = tension; μ = mass per unit length )
Doppler Effect Equation
( fₒ = observed frequency; f𝘴 = frequency of source; c = speed of wave; u = speed of source) Use the - when the source moves towards the observer Use the + when the source moves away from the observer
2: The volume of a television is set so that the sound intensity is 4 x 10^-7 W m^-2. The sound intensity level corresponding to this is 56 dB. Calculate the sound intensity level corresponding to (i) 1 x 10^-7 W m^-2 (ii) 1.6 x 10^-6 W m^-2
(i) 4 x 10^-7 W m^-2 -------- 56 dB 2 x 10^-7 W m^-2 -------- 53 dB 1 x 10^-7 W m^-2 -------- 50 dB (ii) 4 x 10^-7 W m^-2 -------- 56 dB 8 x 10^-7 W m^-2 -------- 59 dB 16 x 10^-7 W m^-2 -------- 62 dB 1.6 x 10^-6 W m^-2 is the same as 16 x 10^-7 W m^-2
A student investigated the variation of the fundamental frequency (f) of a stretched string with its length (l) and obtained the following data. (i) Why was the tension in the string kept constant during the investigation? (ii) Using the above data draw a suitable graph on graph paper to show the relationship between the fundamental frequency of the stretched string and its length. (iii) From the graph estimate the length of the string when its fundamental frequency is 256 Hz l (m) | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | Hz (f) | 675 | 455 | 335 | 273 | 230 | 193 | 173 |
(i) Frequency is dependent on length and tension. To investigate on of these variables, in this case length, the other variable, tension, must be constant (ii) The data has to be adjusted. Get the reciprocal of the values of length and leave the frequency as it is. Hz (f) | 675 | 455 | 335 | 273 | 230 | 193 | 173 | 1/l (1/m) | 5 | 3.33 | 2.5 | 2 | 1.67 | 1.43 | 1.25 | A straight line graph through the origin verifies that frequency is inversely proportional the length. (1/l ∝ f) (iii) when the frequency is 256 Hz the value of the reciprocal length is 1.85. The value of length is given by 1 1.85 = 0.54 m
A person 2 m tall wishes to hang a plane mirror on a vertical wall. What is (i) the minimum length of mirror needed to see himself fully (ii) the distance from the floor to the bottom of the mirror. Assume that the persons eyes are 1.9 m above the floor
(i) Light from the toes strikes the mirror and enters the eyes. Since the angle of incidence equals the angle of reflection the lowest point of the mirror needs to be half way from the ground to the height of the eyes. The distance from the floor to the bottom of the mirror is 1.9 / 2 = 0.95m (ii) Similarly light from the top of the head strikes the mirror and enters the eyes. The top of the mirror is half way between the height of the eyes and the top of the persons head. The top of the mirror is 1.95 m above the floor The total length of mirror needed is 1.95 - 0.95 = 1m , half the persons height.
If the sound intensity level increases by 9 dB, what is the change in the sound intensity
9 dB = increase of 3 dB and increase of 3 dB and increase of 3 dB = double and double and double = 2 x 2 x 2 = 8 times the sound intensity.
Demonstration (laboratory) experiment for the Doppler Effect
A battery powered electronic buzzer is attached to a string 50 cm long. The emitted frequency is set at 300 Hz. The buzzer is spun in circular motion on the end of the string quite fast. An observer standing 2 m away will notice the frequency from the buzzer change as it rotates. The observed frequency changes as the buzzer moves towards and then moves away from the person.
Demonstration (laboratory) of resonance
A number of pendulums are arranged as shown above. Pendulum 1 is made swing in and out of the plane of the page. All the pendulums start to swing a little but pendulum 5 swings most. Pendulums 1 and 5 have the same length and therefore the same natural frequency. Energy is transferred back and forth between the pendulums of the same natural frequency.
What are the two longitudinal standing wave methods
A pipe that is closed at one end A pipe that is open at both ends
An example of a longitudinal stationary wave (use a diagram)
A speaker connected to a signal generator is inserted into the open end of a long graduated cylinder. A layer of lycopodium powder is sprinkled inside the cylinder. Adjust the frequency on the signal generator until a stationary wave is set up. The powder gathers in small bundles at the nodes.
Demonstration experiment to show reflection of sound
A speaker is inserted into a glass tube as shown above. The sound travelling along the glass tube is reflected off the smooth, flat hard surface. The reflected sound is picked up by the microphone and fed to an oscilloscope. Checking the frequency of the display on the oscilloscope verifies it came from the signal generator. The sound absorbing medium reduces the sound energy travelling directly from the speaker to microphone.
Mandatory experiment: To measure the speed of sound in air
A tall graduated cylinder almost filled with water and a hollow glass tube open at both ends are arranged as shown in the diagram. A vibrating tuning fork is placed over the open end of the tube and the tube is raised out of the water until the first resonance between tuning fork and column of air is observed. When resonance happens the frequency of the vibrating column of air equals the frequency of the tuning fork. The resonance in the column of air is detected by hearing, i.e. the sound gets louder. Note and record the frequency that is stamped on the tuning fork. Measure the length of the tube over the water level when first resonance happens and the internal diameter of the glass tube (column of air ) with a sliding callipers. Record the frequencies of several different tuning forks and the corresponding values of the length of the tube ( length of air column ) above the water level. Since you are dealing with the first harmonic for a tube open at one end ( the water acts as a closed end ). Repeat the above calculation for each different frequency of tuning fork and calculate an average value of the speed of sound in air. ƛ/4 = l + d ~> ƛ = 4{l + d} as c = f x ƛ ~> c = f x 4{l + d} ( l = length of tube above water level; d = end correction)
Experiment to measure the focal length of a concave mirror
Adjust the position of the mirror and the screen relative to the light box until the sharpest possible image of the cross-wires is formed on the screen Measure and record the distance from the cross-wires of the light box to the back of the concave mirror, that is the u value. Measure and record the distance from the screen to the back of the concave mirror. That is the v value. Repeat the procedure and record several corresponding values of u and v. Calculate the focal length using 1 / u + 1 / v = 1 / f. Calculate the focal length for each pair of values. Then calculate the average value of f.
Reflection
All objects are visible because light strikes an object or surface and bounces off. Diffuse Reflection and Regular reflection
Accuracy for the experiment to measure the focal length of a concave mirror
Avoid the error of parallax when measuring the image distance and object distance with the metre stick. Avoid small values of u and v as measuring small values results in greater percentage errors. If possible do the experiment in a dark room as your eyes would be more sensitive and better able to determine the sharpest possible image. A matt white screen is best to determine a sharp image. Avoid placing the cross-wires of the light source too near the mirror (inside f ). This would result in a virtual image which cannot be seen on the screen.
How to ensure accuracy with the experiment to investigate the variation of fundamental frequency of a stretched string with length?
Avoid the error of parallax when measuring the length of string with the metre stick Avoid tuning forks of very high frequency. Such tuning forks cause resonance for very small lengths of string. Measuring small lengths results in greater percentage errors. Always ensure the string resonates in the fundamental mode. The paper rider falls off the centre of the string due to an antinode at the centre of the string and nodes at the bridges.
How to ensure accuracy with the experiment to measure the speed of sound in air?
Avoid the error of parallax when measuring the length of the tube ( length of air column ) above the water level Avoid tuning forks of very high frequency as they correspond to small values of length and small values of length result in greater percentage errors. Make sure you always work with the first position of resonance and not the higher harmonics. As you lift the tube the first resonance detected is the first harmonic. The end correction "d" takes account of the gap between the end of the tube and the vibrating tuning fork. The vibrating column of air extends a little beyond the end of the tube. End correction = internal diameter of tube x 0.3
Practical uses of concave and convex mirrors
Concave Mirrors: Shaving mirror or a make-up mirror to give an enlarged image of one's face Dentist's mirror to give an enlarged image of a tooth. Convex Mirrors: Security mirror in a shop to enable staff to see a large area of the shop in one mirror Rear mirror for a car to give a large field of vision
Diffraction (Definition & Diagram)
Diffraction is the spreading out of a wave into the geometrical shadow when it passes through a gap or around an obstacle. Note: Diffraction is much more noticeable when the size of the gap or obstacle is approximately equal to or less than the wavelength of the wave. Sound waves easily diffract as they pass through an open door. The wavelength of the sound wave is approximately equal to the width of the open door. Since light waves have a very small wavelength they require very small gaps to diffract. To demonstrate diffraction of light waves we use a diffraction grating.
Sign conversations for curved mirrors
Distance from mirror to real image is positive Distance from mirror to virtual image is negative Focal length of concave mirror is positive Focal length of convex mirror is negative.
Electromagnetic waves
Electromagnetic waves do not need a medium. They can pass through a medium but they travel faster in a vacuum than in a medium. They consist of varying electric and magnetic fields. Examples are light, radio waves, X-rays,
Coherent sources
Emit waves of the same frequency and the waves are in phase or have a constant phase difference.
How do you reduce noise using destructive interference?
Example: You are in a room with a lot of noise coming in from the street. A microphone captures some of this noise, feeds it to an electronic device to analyse the wave pattern. The wave pattern is inverted and emitted as a sound from the device. Your ears now receive two sounds which are the exact "opposite" of each other. Due to destructive interference you hear no sound.
The threshold of hearing (Definition)
For a frequency of 1,000 Hz the threshold of hearing is the lowest sound intensity the human ear is capable of hearing
Frequency
Frequency (f) is the number of complete oscillations made by a point in one second. Unit is the Hertz ( Hz )
Harmonics
Harmonics are multiples of the fundamental mode of vibration The overtones are all the harmonics except the first harmonic.
Ray Diagram Rules
If a ray of light travels parallel to the principle axis heading for the mirror it reflects through the focal point If a ray of light travels through the focal point heading for the mirror it reflects parallel to the principal axis. If a ray of light strikes the mirror at the pole then the angle of incidence equals the angle of reflection. If a ray of light travels through the centre of curvature heading for the mirror it reflects back along its own path
Fundamental mode of vibration
If a string is fixed at both ends and plucked in the centre the most simple or fundamental type of vibration. For the fundamental mode the string vibrates at its lowest frequency
Curved Mirrors
Imagine a piece of glass being cut from a hollow glass sphere. If you silver the outside of the piece of glass you get a concave mirror If you silver the inside of the piece of glass you get a convex mirror.
Compressions and Rarefactions.
In Longitudinal waves, the energy (from the vibrating tuning fork) passes through the surrounding air. The regions of high particle density ( high air molecule density) are called compressions and the regions of low particle density are called rarefactions.
Interference (Definition & Diagram)
Interference will occur when waves from coherent sources meet and combine. Constructive interference gives a wave of greater amplitude. Destructive interference gives a wave of smaller amplitude.
Harmonics of pipes that are closed at one end
It can only produce odd harmonics
Pipes that are open at both ends
It is possible to make the air vibrate in a pipe open at both ends. The air can vibrate at certain frequencies and these are called the harmonics There is an antinode at both open ends All the harmonics are possible. Musical instruments based on this are called good quality instruments since all the harmonics are possible. Examples are : tin whistle, flute, pan pipes
Pipes that are closed at one end
It is possible to make the air vibrate in a pipe which is closed at one end and open at the other end. The air can vibrate at certain frequencies and these are called the harmonics. There will be a node at the closed end and an antinode at the open end. You will notice that only the odd harmonics are possible for the pipe closed at one end. Musical instruments based on this are called poor quality instruments since only the odd harmonics are available. Examples are : clarinet, trombone, saxophone
Laws of Reflection
Law 1: The incident ray, the reflected ray and the normal all lie in the same plane. Law 2: the angle of incidence = the angle of reflection i = r
Applications of the Doppler Effect
Measuring the speed of a star Checking the speed of a car: An electromagnetic wave of known frequency f𝘴 is directed at a moving car. The speed of this wave is also known as all electromagnetic waves the same speed. The value of the speed of the wave c is 3 x 10^8 m/s The wave reflected off the car will have a different frequency to the incident wave. This reflected frequency fₒ is measured electronically. The values of f𝘴, fₒ and c are now all known. We calculate the speed of the car (u) using the Doppler Effect Equation
Mechanical waves
Mechanical waves need a medium They are made by a disturbance in the medium They pass along a medium by causing the particles of the medium to vibrate This vibration is passed from one particle to its neighbour. The overall motion of the particles of the medium is zero Examples are sound waves, ultrasound waves, waves on water
In an experiment to measure the value of the speed of sound in air a student recorded the frequency f of several tuning forks and the corresponding values of the length of tube l above the water level. The internal diameter of the tube was 3 cm. Calculate the value of the speed of sound in air. Hz (f) | 256 | 288 | 320 | 341 | 384 | 480 | 512 | l (cm) | 32 | 29 | 26 | 24 | 21 | 17 | 16 |
Method 1: the end correction = 3 x 0.3 = 0.9 cm Hz (f) | 256 | 288 | 320 | 341 | 384 | 480 | 512 | l (cm) | 32 | 29 | 26 | 24 | 21 | 17 | 16 | l + d (cm) |32.9 | 29.9 |26.9 | 24.9 | 21.9 | 17.9 | 16.9 | ƛ = 4{l+d}(m)| 1.316| 1.196 |1.076 | 0.996 |0.876 |0.716|0.676| c = fx4{l + d} (m/s) | 337| 344 |344 | 340 |336 |344|346| The average of the seven values is c = 341.6 m/s Method 2: Plot a graph of frequency on the y-axis and reciprocal wavelength on the x-axis. Hz (f) | 256 | 288 | 320 | 341 | 384 | 480 | 512 | l (cm) | 32 | 29 | 26 | 24 | 21 | 17 | 16 | l + d (cm) |32.9 | 29.9 |26.9 | 24.9 | 21.9 | 17.9 | 16.9 | ƛ = 4{l+d}(m)| 1.316| 1.196 |1.076 | 0.996 |0.876 |0.716|0.676| 1/ƛ (1/m) | 0.76 | 0.84 |0.93 | 1.00 | 1.14 | 1.40 | 1.48 | Speed = c = f/(1/ƛ ) = y2-y1/x2-x1 =430 - 170/1.25 - 0.5 = 347 m/s
The household mirror ( Plane mirror )
Most household mirrors consist of a sheet of glass with a reflecting layer (silvering) on one side. The sheet of glass refracts (bends) the light causing an unnecessary complication. To simplify matters we usually ignore the sheet of glass and consider the plane mirror as just the reflecting layer
Natural Frequency vs Forced Frequency
Natural frequency is when a stretched elastic band is plucked in the middle it will vibrate at its natural frequency. Forced frequency is when an external vibration force acts on a system that is capable of vibrating the external force provides the forced frequency. Touch a vibrating tuning fork to a stretched string. The string vibrates at the same frequency as the tuning fork. The frequency of the tuning fork is the forced frequency.
Nodes and Antinodes
Nodes are points along the medium where the displacement is zero Antinodes are points along the medium where the displacement is a maximum
Pitch
Pitch is related to the frequency of a sound. The higher the frequency the higher the pitch "pitch is to sound as colour is to light"
An object is placed 16 cm in front of a concave mirror and an inverted image is formed 24 cm in front of the mirror. The object is 2 cm in size. Calculate the focal length of the mirror and the size of the image.
Real Image: 1 / u + 1 / v = 1 / f 1 / u + 1 / v = 1 / f 1 / 16 + 1 / 24 = 1 / f 3 + 2 / 48 = 1 / f 5f = 48 f = 9.6 cm Magnification = image size / object size = v / u = image size / 2 = 24/16 16 x image size = 48 image size = 3 cm
diffuse reflection
Reflection that occurs when parallel rays of light hit a rough surface and all reflect at different angles An example would be light from the paper the notes are written on
Resonance
Resonance is the transfer of energy between two bodies of the same natural frequency It happens when a vibrating system responds with maximum amplitude to a forced frequency. An example would be a singer who shatters a wine glass. The wine glass is the system capable of vibrating. It has a natural frequency. The sound from the singer provides the forced frequency. When forced frequency equals the natural frequency resonance happens. The glass shatters.
Mandatory Experiment: To investigate the variation of fundamental frequency of a stretched string with tension
Select a suitable length of string and keep this length constant, place a small paper rider on the string Note and record the frequency of the first tuning fork. Now let the base of this vibrating tuning fork touch the string, adjust the weight in the pan (adjust the tension) until the string gives resonance in the fundamental frequency, i.e. the paper rider is thrown from the string. Note and record the tension for this first position of resonance. Note and record several values of tuning fork frequencies and the corresponding values of tension that caused resonance, plot a graph of frequency on the y-axis and square root of tension on the x-axis. A straight line graph through the origin verifies that frequency is proportional to the square root of the tension
Sound Intensity (Definition & Equation)
Sound intensity is the sound energy per second per unit area. Sound intensity = sound energy per sec / area energy passes through = joules (/s) / m^2 = watt m^-2
Decibel adapted scale ( dBA )
Sound level meters use special filters to emphasise sounds with frequencies values between 2 000 Hz and 4 000 Hz. Higher and lower frequencies are attenuated ( partially blocked out ). The human ear is most sensitive to sounds within this frequency range. These frequencies can cause most damage to the human ear. They cause resonance in the human ear canal. The decibel adapted scale takes into account the fact that the human ear is more sensitive to certain frequencies of sound.
Stationary ( Standing ) Waves (Definition & Diagram)
Stationary waves are waves where there is no net transfer of energy They are produced when two progressive waves of the same amplitude, same frequency, same speed and moving in opposite directions meet. Stationary waves are produced on the strings of musical instruments and in the air columns of wind instruments. The example above of a stationary wave on a string would be a transverse stationary wave.
A string 70 cm long has a fundamental frequency of 256 Hz when subjected to a tension of 45 N. Calculate the mass per unit length of the string.
Step 1: Square both sides Step 2: Rearrange Formula μ = 3.5 x 10^-4 kg/m
Mandatory Experiment: To investigate the variation of fundamental frequency of a stretched string with length (with Diagram)
Subject the string to a fixed tension and keep this tension constant, place a small paper rider on the string. Note and record the frequency of the first tuning fork. Now let the base of this vibrating tuning fork touch the string, adjust the moveable bridge until the string gives resonance in the fundamental frequency, i.e. the paper rider is thrown off the string. Note and record the length of the string for this first position of resonance. Note and record several values of tuning fork frequencies and the corresponding lengths of string when resonance happens, plot a graph of frequency on the y-axis and the reciprocal of length on the x-axis. A straight line graph through the origin verifies that frequency is inversely proportional the length.
The Doppler Effect (Definition & Diagram)
The Doppler effect is the apparent change in the frequency of a wave due to the relative motion between source and observer. The observed wavelength and frequency change as the source moves past observer.
The table shows the measurements recorded by the student for the object distance u and the image distance v. u/cm | 20.0 | 25.0 | 35.0 | 45.0 | v/cm | 66.4 | 40.6 | 27.6 | 23.2 | Using the data in the table, find an average value for the focal length of the mirror.
The average value of focal length is 15.4 cm
Characteristics of a sound:
The characteristics of a sound are pitch, quality and loudness. These are the three ways in which sounds differ.
Condition for constructive and destructive interference
The condition for constructive interference is that when the waves meet they are in phase and the waves have the same frequency. The crest of one wave meets the crest of the other wave. The condition for destructive interference is that when the waves meet they are out of phase with a path difference of ƛ/2 or a multiple of this. The crest of one wave meets the trough of the other wave. Waves are in phase when they are doing the same thing at the same time.
Longitudinal Wave (Definition & Diagram)
The direction in which the energy travels is parallel to the direction of vibration of the particles of the medium.
Transverse Wave (Definition & Diagram)
The direction in which the energy travels is perpendicular to the direction of vibration of the particles of the medium.
Explanation of the demonstration experiment to show the wave nature of sound
The distances from the speakers to the central position of loudness (A) are equal. If the waves leave the speakers in phase then they arrive at A in phase and constructive interference occurs i.e. the sound is loud. The distances from the speakers to the first positions of faint sound (B) and (C) are not equal. There is a difference of ƛ/2 in the distances. Thus waves that left the speakers in phase will arrive at B and C out of phase. Destructive interference will occur and the sound is faint. The distances from the speakers to the first positions of loudness (D) and (E) are not the same. There is a difference of ƛ in the distances. Waves that left the speakers in phase will again be in phase when the arrive at D and E. Constructive interference will happen and a loud sound is heard. The distances from the speakers to the second positions of faint sound (F) and (G) are not the same. There is a difference of 3ƛ/2 in the distances. Waves that left the speakers in phase will arrive at F and G out of phase. Destructive interference will occur and the sound is faint.
The frequency range of audibility (Definition)
The human ear is only able to hear certain frequencies. The frequency range of audibility for the human ear is between 20 Hz and 20 000 Hz. 20Hz < f < 20000 Hz
The Human Ear
The human ear is only able to hear certain frequencies. The frequency range of audibility for the human ear is between 20 Hz and 20 000 Hz. The human ear can hear sound intensities between 1 x 10^-12 Wm^-1 (threshold of hearing) and 1 Wm^-1 (threshold of pain). We cannot hear sound intensities below the threshold of hearing, and above the threshold of pain can permanently damage our hearing The upper and lower limits for sound intensity can vary depending on the frequency of the sound. The values quoted above assume a frequency of 1,000 Hz
Sound Intensity Level
The human ear responds to changes in sound intensity in a rather unique way, if the power of a sound increases from 0.1 W to 1 W the change is the same as if the power increases from 1 W to 10 W. The ear responds to the ratio of the powers not their difference. Sound intensity level which is measured in decibels due to this. Sounds are compared to a standard which is 1 x 10^-12 Wm^-1 If the sound intensity is doubled the sound intensity level increases by 3 dB If the sound intensity is halved the sound intensity level decreases by 3 dB
Ray Diagram - the source of light (object) is positioned very far (at infinity ) from the concave mirror.
The image is formed on the focal point. The image is a real image
Ray Diagram - the source of light (object) is positioned on f .
The image is located at infinity. The image is real, inverted and greatly magnified. v = + u = + f = +
Ray Diagram - the source of light (object) is positioned anywhere in front of the convex mirror
The image is located behind the mirror, inside f . The image is virtual, erect and diminished. v = - u = + f = -
Ray Diagram - the source of light (object) is positioned inside f .
The image is located behind the mirror. The image is virtual, erect and magnified. v = - u = + f = +
Ray Diagram - the source of light (object) is positioned a little outside c.
The image is located between c and f. The image is real, inverted and diminished. v = + u = + f = +
Ray Diagram - the source of light (object) is positioned on c .
The image is located on c. The image is real, inverted and same size as object. v = + u = + f = +
Ray Diagram - the source of light (object) is positioned between c and f.
The image is located outside c. The image is real, inverted and magnified. v = + u = + f = +
Image in a plane mirror
The image of an object in a plane mirror is the same distance behind the mirror as the object is in front of the mirror. u = v (u = distance from object to mirror; v = distance from image to mirror.) The image in a plane mirror is a virtual image. Rays of light do not actually pass through a virtual image, they only appear to do so. The image in a plane mirror is laterally inverted. The left side appears to be the right side and vice versa.
Regular reflection
The incident beam of parallel light is reflected as a beam of parallel light. An example would be light reflected from a highly polished surface.
Loudness
The loudness of a sound depends on the amplitude of the vibration of the sound wave. Loudness is subjective.
Amplitude
The maximum displacement of a particle of the medium from its mean position.
Factors affecting the natural frequency of a stretched string
The natural frequency of a stretched string depends on: The length of the string The tension force the string is subjected to. The mass per unit length of the string. (mass of 1m of string)
Quality
The same note played on two different instruments does not sound the same e.g. the piano and the violin. The quality of a sound depends on the number and intensity of the harmonics (overtones) present
Sound is emitted from a source at a rate of 2.6 Watt. Assuming that the sound is emitted equally in all directions and that no sound is absorbed or reflected, calculate the sound intensity at points 8.6 m from the source.
The sound energy spreads out from the source like an expanding sphere. The area used for the calculation will be the surface area of a sphere of radius 8.6 m. Sound intensity = sound energy per sec / area energy passes through = power / 4πR^2 = 2.6 / 4π x 8.6^2 = 2.8 x 10^-3 Wm^-2
Demonstration experiment to show refraction of sound
The speaker connected to the signal generator emits sound in all directions. Some of this sound is refracted by the balloon of CO2 gas and is concentrated at a point. This sound energy is picked up by the microphone and displayed on the oscilloscope.
How to ensure accuracy with the experiment to investigate the variation of fundamental frequency of a stretched string with tension?
The tension is equal to the weight of the pan and its contents. Avoid small values of frequency as they correspond to small values of weight. These small values lead to greater percentage errors. Always ensure the string resonates in the fundamental mode. The paper rider falls off the centre of the string due to an antinode at the centre of the string and nodes at the bridges.
Demonstration experiment to show the wave nature of sound
The two speakers are connected in parallel to a signal generator and emit sounds of the same frequency and amplitude and are in phase. The speakers are coherent sources. A person walking along the line X Y will notice the sound intensity vary from loud to faint in a regular manner. Constructive and destructive interference is taking place. Interference is a phenomenon associated with waves, so if sound exhibits interference then we conclude that sound is a wave.
Solving numerical problems on mirrors
There are two equations: 1 / u + 1 / v = 1 / f Magnification = image size / object size = v / u (f = focal length; u = object distance; v = image distance)
Progressive waves
These waves consist of energy moving away from a source. Energy is transferred from one place to another. An example of this is that the students in this room can hear the sweet tones of my beautiful voice as the sound waves spread outwards from my vocal chords Progressive waves are different to the standing waves
Different decibal rankings
Threshold of hearing 0 dB Whispering 30 dB Normal conversation 60 dB Noisy factory 90 dB Loud thunderclap 110 dB Threshold of pain 120 dB
Demonstrate the difference between transverse and longitudinal waves
Transverse waves can be polarised but longitudinal waves cannot be polarised.
Wavelength
Wavelength (ƛ) is the distance between two successive crests or two successive compressions. Unit is the metre
Forced vibration
When a vibrating tuning fork touches a table top the sound from the tuning fork appears much louder. The table top is forced to vibrate at the same frequency as the tuning fork.
Demonstration experiment to show that sound requires a medium in which to travel
When the switch is closes you will see the bell is ringing and you will hear the sound from the bell. As the air is removed from the bell jar you will notice that the sound becomes more faint. When all the air is removed you will hear no sound. You will see that the electric bell is still working. We conclude that sound cannot travel in a vacuum. Sound requires a medium in which to travel.
Wavefronts
When waves spread out from a source are often represented by the diagram below. The source refers to the origin of the wave. A line which shows the direction of travel of the wave is called a ray. The concentric circles above are called wavefronts. The distance between consecutive wavefronts is equal to the wavelength.
Harmonics of pipes that are open at both ends
all harmonics are possible
wavelength and frequency equation
c = f x ƛ
Calculate the fundamental frequency of a string 80 cm long, subjected to a tension of 22 N if the mass per unit length is 4 x 10^-4 kg/m
f = 146.6 Hz
A convex mirror of focal length 20 cm forms an image that is one quarter the size of the object. Find the value of the object distance and the image distance.
magnification = 1 / 4 = v /u therefore u = 4v Virtual Image: 1 / u - 1 / v = - 1 / f 1 / 4v - 1 / v = - 1 / 20 1 - 4 / 4v = - 1 / 20 - 4v = - 60 v = 15 cm As u = 4v ~> u = 60 cm
An image formed in a concave mirror of focal length 12 cm is three times the size of the object. Calculate the positions of the object for this to happen.
magnification = 3 = u / v ~> v = 3u Real Image: 1 / u + 1 / v = 1 / f 1 / u + 1 / 3u = 1 / 12 3 + 1 / 3u = 1 / 12 3u = 48 u = 16 cm object distance = u = 16 cm Virtual Image: 1 / u - 1 / v = 1 / f 1 / u - 1 / 3u = 1 / 12 3 - 1 / 3u = 1 / 12 3u = 24 u = 8 cm object distance = u = 8 cm
Terminology for curved mirrors
p is the pole of the mirror. The pole of the mirror is the centre of the mirror itself c is the centre of curvature of the mirror. It is the centre of the sphere from which the mirror was cut. f is the focal point of the mirror. It is half way between the pole and the centre of curvature. The distance between p and f is called the focal length. The symbol for the focal length is the letter f. The straight line that joins the points c , f , and p is called the principal axis.
A radio station broadcasts at 92 M Hz. Calculate the corresponding value of wavelength. (c = 3 x 10^8 m/s)
rearrange c = f x ƛ to give ƛ = c / f ƛ = (3 x 10^8) / (92 x 10^6) ƛ = 3.26m
Given that the speed of sound in air is 340m/s, calculate the wavelength of a musical note of frequency 256 Hz.
rearrange c = f x ƛ to give ƛ = c / f ƛ = 340 / 256 ƛ = 1.328m
A whistle emitting a note of 1 kHz is whirled in a horizontal circle on the end of a string 1.2 m long at a constant angular speed of 50 rad/s. What are the highest and lowest frequencies heard by a person standing some distance away? (speed of sound in air =340m/s)
step 1: change angular speed to linear speed v = r x ω v = 1.2 x 50 v = 60 m/s
The relationship between the natural frequency of a stretched string and its tension was investigated by a student. The length of the string was 64 cm. The following values of frequency f and tension T were obtained. By drawing a suitable graph on graph paper calculate (i) the frequency when the tension was 16 N Hz (f) | 256 | 288 | 320 | 384 | 427 | 480 | 512 | T (N) | 5 | 6 | 9 | 12 | 14 | 18 | 22 |
the data has to be adjusted, get the square root of all the tension values but leave the frequency values as they are. Hz (f) | 256 | 288 | 320 | 384 | 427 | 480 | 512 | √t (N¹'²) | 2.2 | 2.4 | 3 | 3.5 | 3.7 | 4.2 | 4.7 | (i) When the tension is 16 N the square root of tension is 4. Now 4 on the x-axis corresponds to 450 Hz on the y-axis.
A wave of frequency 2400 Hz is reflected off a surface back along its own path so that a stationary wave is set up. The distance between successive nodes is 7 cm. What is the speed of the wave?
ƛ = 2 distance between successive nodes = 2 x 7 = 14 cm 14cm = 0.14m c = f x ƛ c = 2400 x 0.14 c = 336m/s
A pipe closed at one end is 30 cm long. The air in the pipe is made vibrate at the fundamental frequency (1st harmonic). Calculate the value of this fundamental frequency. (speed of sound in air = 340 m/s)
ƛ/4 = 30cm therefore ƛ = 120cm = 1.2m f = c/ƛ f = 340/1.2 f = 283.3 Hz
