| |
There are three types of waves:
- 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.
- Electromagnetic waves (e.g. light, radio) do not require
a medium (air, water) to travel in.
- 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
|
|
