WAVES - LCHS Physics LCHS Main SiteSD271 Main site
Objectives: The learner will...
...study the principles of waves and solve problems including: frequency, period, wavelength, velocity, interference and the Doppler effect
 


by Stephanie Rasmussen

There are three types of waves:

  1. Mechanical waves require a material medium to travel (air, water, ropes). These waves are again divided into three different sub-types. 1)Transverse waves cause the medium to move perpendicular to the direction of the wave. 2) Longitudinal waves cause the medium to move parallel to the direction of the wave. 3) Surface waves are both transverse waves and longitudinal waves mixed in one medium.
  2. Electromagnetic waves (e.g. light, radio) do not require a medium (air, water) to travel in.
  3. Matter waves are produced by electrons and particles.

The length of a wave (), or wavelength is measured in meters from crest to crest or trough to trough. The number of waves that pass a fixed point in 1 second is called the frequency ( f ) and is measured in Hertz ( Hz, waves/cycles per second ). The period ( T ) of a wave is the time (seconds) for one wave to pass a fixed point, or the inverse of the frequency ( f = 1 / T ). The velocity of the wave ( v ) is a product of the frequency and the wavelength. ( v =f )


Problem 1 : A wave with a period of .16666 secs has a frequency of _ Hz
The frequency is the inverse of the period and vice versa.
So, f = 1 / .16666 = 6 Hz

Problem 2 : A .54 m wave with a frequency of 87 Hz, has a velocity of _ m/s
The velocity of a wave is its wavelength times its frequency.
So, v =f = .54 x 87 = 46.98 m/s

Other examples of periodic/wavelike motion include pendulums:

Problem 3 : A _ m long pendulum has a period of 4.2 secs
A simple pendulum has a period (T) equal to 2pi x the square root of [its length divided by the acceleration due to gravity]
So, 4.2 = 2 x sqrt (L / 9.8) ...solving for L, L = 9.8(4.2/
2)2 = 4.3789 meters
Problem 4 : A 3.2 kg pendulum with I = 4 and .26 m from pivot to center mass has a frequency of _Hz
An irregular shaped pendulum with a known moment of Inertia has a period T = 2pi times the square root of [its Inertia divided by (its mass x acceleration due to gravity x its Length from pivot to center mass)]. Once you find the period (T) just take the inverse to find its frequency.
So, T = 2 x sqrt (4 / (3.2 x 9.8 x .26)) = 4.4 sec, and f = 1 / 4.4 or .227 Hz

The general wave equation which incorporates amplitude, frequency and time is:
y = A sin ( 2 f t )
A= amplitude, f = frequency, t = time (seconds) Be sure your calculator is in Radian Mode for this equation to work!!! This equation can be used to give/plot the value of a wave as time passes. The value oscillates between positive and negative values of the amplitude and follows the sinousoidal (wavelike) shape.
Problem 5 : A wave with peak amplitude 80 has an amplitude of _ after .005 s with a f = 42 Hz
The instantaneous amplitude at an exact time for a wave equals the amplitude times the sine of (2pi times the frequency times the time). Be sure your calculator is in Radian Mode for this equation to work!
So, y = 80 sin(2 x 42 x .005) = 77.48665

2 different waves can be added together giving constructive and destructive interference effects giving results can be positive, negative or even zero.
y = A1 sin ( 2 f1 t1 ) + A2 sin ( 2 f2 t2 )

Problem 6 : Wave A of amp 23 and f 58 Hz combines with wave B of amp 34 and same f
have a combined amplitude of _ after .0052 secs

The instantaneous amplitude at an exact time for 2 different waves is just like the previous problem. Do for each wave and add the individual y's together. It is possible to get a negative result or even zero!
So, y = 23 sin ( 2 x 58 x .0052 ) + 34 sin ( 2 x 58 x .0052 )
this can simplify to y = ( 23 + 34 ) sin ( 2 x 58 x .0052 ) = 54.03

Problem 7 : Wave A with amp 50 and f 60 Hz starts behind wave B by .0048 s
Wave B has the same amp and f. Their combined amp after .0012 s = _

The instantaneous amplitude at an exact time for 2 different waves is just like the previous problems. Do for each wave and add the individual y's together. HINT: t2 = .0048 + .0012 = .006 seconds.
So, y = 50 sin ( 2 60 x .0012 ) + 50 sin ( 2 60 x .006 ) = 60.38

Problem 8 : Wave A of amp 32 and f 87 Hz combines with wave B with same amp and f 66.
Their combined amp after .0023 seconds is _

The instantaneous amplitude at an exact time for 2 different waves is just like the previous problems. Do for each wave and add the individual y's together.
So, y = 32 sin ( 2 87 x .0023 ) + 32 sin ( 2 66 x .0023 ) = 56.54

Problem 9 : Wave A with amp 51 and f 33 Hz combines with wave B with amp 80 and f 24 Hz.
Their combined amp after .0011 seconds is _

The instantaneous amplitude at an exact time for 2 different waves is just like the previous problems. Do for each wave and add the individual y's together.
So, y = 51 sin ( 2 33 x .0011 ) + 80 sin ( 2 24 x .0011 ) = 24.74

The DOPPLER EFFECT is a change in frequency due to the source of the wave moving with a rapid velocity causing a scrunching or stretching of the wave. The sudden increase and then decrease of pitch in the sound of a passing race car is an example. A star travelling toward us in the universe causes a spectral shift toward the blue while away causes a red shift. Click Here for a great Doppler Effect Demo
f = fs (v + vr)/(v - vs ) ...this equation for sound waves
fs = source frequency, vr = receiver velocity (might be zero), vs = source velocity
v (for sound) is about 346 m/s at room temperature

Problem 10 : If the speed of sound is 346 m/s, a 880 Hz sound source
traveling toward you at 10.2 m/s has a perceived f of _ Hz

The speed of sound is about 346 m/s at room temperature, source velocity (Vs) is 10.2 m/s, receiver velocity (Vr) is zero, and the source frequency (fs) is 880.
So, f = 880 (346 + 0)/(346 - 10.2 ) = 906.73 Hz


 
Sample Problems
PB1
A wave with a period of .08333334 secs has a frequency of _ Hz
PB2
A .71 m wave with a frequency of 67 Hz, has a velocity of _ m/s
PB3
A _ m long pendulum has a period of 8.3 secs
PB4
A 5.9 kg pendulum with I = 28 and .86 m from pivot to center mass has a frequency of _Hz
PB5
A wave with peak amplitude 88 has an amplitude of _ after .02 s with a f = 55 Hz
PB6
Wave A of amp 82 and f 27 Hz combines with wave B of amp 34 and same f
have a combined amplitude of _ after .14 secs
PB7
Wave A with amp 40 and f 62 Hz starts behind wave B by .56 s
Wave B has the same amp and f. Their combined amp after .17 s = _
PB8
Wave A of amp 21 and f 84 Hz combines with wave B with same amp and f 32.
Their combined amp after .23 seconds is _
PB9
Wave A with amp 79 and f 47 Hz combines with wave B with amp 83 and f 26 Hz.
Their combined amp after .09 seconds is _
PB10
If the speed of sound is 343 m/s, a 2000 Hz sound source
traveling toward you at 5.4 m/s has a perceived f of _ Hz
answer bank:

35.182
17.10104
-113.9453
13.93

29.9736
2031.709
12
.21209449
47.57
148.4563
.2131739
51.7251022


Resources (Study Links/Study Tips/Reading Lists)

Wave addition demo program
Doppler Effect Lab

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