Advice & Insight From BMAT Specialists
The diagram shows a typical wave.
CREST: The highest point on a wave.
TROUGH: The lowest point on a wave.
AMPLITUDE (A): The maximum displacement of the wave from the undisturbed position. (Measured in metres/cm).
WAVELENGTH (λ): The distance between two adjacent corresponding points. (Measured in metres/cm).
FREQUENCY (f): The number of oscillations/vibrations/cycles per second. (Measured in Hertz (Hz). [ 1 Hz = 1 s⁻¹]
PERIOD (T): The time for one vibration/oscillation/cycle of the wave. (Measured in seconds).
CYCLE: One complete ‘up and down’ part of a wave.
From the definitions it follows that:
f = 1/T and T = 1/f
[It is important to note that T MUST be in seconds for f to be in Hz]
Example 1
Calculate the frequency of a wave if the period is:
(i) 5s __________ (ii) 0.02s _____________ (iii) 1 ms _____________ (iv) 20ms ____________ (v) 10μs ____________
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Example 2
Calculate the period of a wave if the frequency is:
(i) 4Hz ____________ (ii) 0.02Hz ___________ (iii) 1kHz ___________ (iv) 2 MHz ____________ (v) 0.5MHz __________
It can be show that for ANY wave: v = f λ
[ CARE! Do NOT mix the units for v and λ. They must BOTH be in metres or BOTH in cm].
ALSO, note that the speed v of any wave is also given by: speed (v) = distance
time
Learn time-efficient BMAT strategies and practice with reflective BMAT questions & worked solutions.
Example 3
Calculate the speed of wave if:
(i) the frequency is 0.1kHz and wavelength 10 cm
(ii) the frequency is 0.2 MHz and the wavelength is 100μm
Example 4
Calculate the frequency of a wave if:
(i) the speed is 100 ms⁻¹ and the wavelength is 0.5 m,
(ii) the speed is 3 x10⁸ ms⁻¹ and the wavelength is 1.5μm.
Example 5
Calculate the wavelength of a wave if:
(i) the speed is 0.2 ms-1 and the frequency is 0.5 Hz,
(ii) the speed is 25 ms-1 and the frequency is 0.5 kHz.
If the particles of a wave vibrate at right angles to the wave direction the wave is said to be transverse.
Examples of transverse waves: ALL em waves (e.g. light, radio, X-rays…)
A TRANSVERSE wave can easily be shown by a slinky or rope as shown below.
If the particles of a wave vibrate parallel to the wave direction the wave is said to be longitudinal.
Example of longitudinal wave: Sound.
A LONGITUDINAL wave can best be shown by a slinky as shown below.
The centre of a ‘squashed’ region is called a compression (C).
The centre of a ‘stretched’ region is called a rarefaction (R).
Example 6
The diagrams show a progressive wave. Displacement d is plotted on the vertical axis and either displacement (x) or time (t) is plotted on the horizontal axis. The period (T), wavelength (l) and amplitude (A) are also shown.
However, only one graph is correctly labelled.
Which graph is correctly labelled?
[ANSWER: D]
The diagrams below refer to the same wave.
(a) Indicate on the relevant graph the amplitude, period and wavelength.
(b) Use the graphs to calculate the speed of the wave.
When an object e.g. a tuning fork vibrates it causes the surrounding air molecules to vibrate. These vibrations are passed on through the air. When the vibrations reach the ear drum it also is forced to vibrate at the same frequency. These vibrations are converted into an electrical signal to the brain and a ‘sound’ is heard.
If there was no air between the source (tuning fork) and the person then NO vibrations can be passed on. So it follows that sound needs a medium (solid, liquid or gas) to travel.
Sound travels fastest through a solid and slowest through a gas (air).
The human ear can detect sounds with frequencies from about 20 Hz to about 20 kHz.
The frequency of a sound wave is sometimes called the pitch.
The greater the amplitude of a sound wave the louder the sound.
Sound waves are longitudinal waves and can have wavelengths of about 20 cm.
When a sound wave is reflected from a surface an echo is formed.
Ultrasound are sound waves with frequencies greater then 20 kHz.
Typical values are in the MHz range.
Ultrasound can be used for sonar, medical imaging of body organs and measuring blood flow.
Suppose the source moves at a constant speed towards observer A. The crests of the wave from the source are squashed together in the direction of A and spread apart in the direction of B.
This is shown in the second diagram.
The waves reaching observer A are arriving with a shorter wavelength and hence a higher frequency (remember the speed of the waves is constant).
The waves reaching observer B are arriving with a longer wavelength and hence a lower frequency.
Both observers hear sounds of different frequencies.
Person A will hear a higher frequency than f. Person B will hear a lower frequency than f.
The faster the source moves the greater the change in frequency detected. In this way the speed of a car or flow of blood can be measured.
This change in frequency (and hence wavelength) is called the Doppler effect.
(Small wavelength—LARGE FREQUENCY)
(LARGE WAVELENGTH – small frequency)
General Properties
REFLECTION
The diagram shows a light ray incident on a plane mirror and being reflected.
The dotted line is called a normal and is at an angle of 900 to the mirror.
The angle between the incident ray and the normal is called the angle of incidence (i).
The angle between the reflected ray and the normal is called the angle of reflection (r).
Experiment shows that : i = r
[Note: the speed (v), frequency (f) and wavelength (λ) do NOT change during reflection]
REFRACTION
When light travels from one medium (e.g. air) into another medium (e.g. water or glass) it changes direction.
This change in direction is called refraction.
This occurs because when light enters a different medium its speed changes.
If the speed decreases (e.g. light passing from air to glass/water) the light ray bends towards the normal.
If the speed increases (e.g. light passing from glass/water to air) the light bends away from the normal.
The angle between the refracted ray and the normal is called the angle of refraction (r).
Note:
Although they all travel at the SAME speed in air, they travel at DIFFERENT speeds in glass/water.
Example 1:
(i) 5s f = 1/T = 1/5 = 0.2Hz (ii) 0.02s f = 50Hz (iii) 1 ms f = 1000Hz (iv) 20ms f = 50Hz (v) 10μs f = 10⁵ Hz
Example 2:
(i) 4Hz T = 1/f =1/4 = 0.25s (ii) 0.02Hz T= 50s (iii) 1kHz T = 1 ms (iv) 2 MHz T = 0.5μs (v) 0.5MHz T = 2μs
Example 3:
(i) the frequency is 0.1kHz and wavelength 10 cm
v = fλ = 100×0.1 = 10 ms⁻¹
(ii) the frequency is 0.2 MHz and the wavelength is 100μm
v = 2 x 10⁵ x100 x 10⁻⁶ = 20 ms⁻¹
Example 4:
(i) f = v/λ = 100/0.5 = 200Hz
(ii) f = (3×10⁸)/(1.5×10⁻⁶) = 2×10¹⁴Hz
Example 5:
(i) λ = v/f = 0.2/0.5 =0.4m
(ii) λ = 25/(500) = 0.05m
Example 6:
(a) Indicate on the relevant graph the amplitude, period and wavelength.
Either graph for Amplitude, 1st graph for λ, 2nd graph for Period.
(b) Use the graphs to calculate the speed of the wave.
From 1st graph λ ≈ 3.4m From 2nd graph T = 0.2s f = 5HZ v = f λ = 5×3.4 = 17 ms⁻¹
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