Lecture 5
Motion of particles in
spring is parallel or
anti-parallel to velocity
of wave
Motion of particles in rope
is perpendicular to velocity
Transverse Wave
A Longitudinal Wave
y(x,t)=ym sin( kx- 7t) = 3 sin(2x -2t)
Example
Wavelength and Frequency
y(x,0) = 3 sin(2x)
• y=f(x,t) = f(x-vt) gives the shape of the
wave at a given time (i.e. snapshot)
• for SHM of one end of the string, the
function is a sine or cosine
• y(x,t) = ym sin(kx - 7t) “travelling wave”
• ym is the “amplitude”
• (kx - 7t) is the “phase”
Repeats in space with period wavelength
T
Focus on point x=0
y(0,t) = 3 sin(-2t)
Repeats in time with period T
Wavelength and Frequency
y(x,t)=ym sin( kx- 7t)
Wavelength and Frequency
• y(x,t)=ym sin( kx- 7t)
• y(x+,t)= ym sin( kx- 7t+k )= y(x,t)
is the ‘wavelength’
• hence k =2%
• angular wave number k= 2%/ rads/m
Snapshot at t=0
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Consider point x=0
y(0,t)=ym sin(- 7t) = - ym sin(7t)
y(0,t+T)= -ym sin( 7t+ 7T)=y(0,t)
hence 7T= 2%
angular frequency 7=2%/T rads/s
=2%f
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Consider two snapshots of the wave taken t apart
Speed
• y(x,t)=ym sin( kx- 7t)
• rewrite as y(x,t)=ym sin[ k(x- (7/k)t)]
• general form of wave travelling to
the right is f(x-vt)
• hence v = 7/k
•
7= v k
1 2
The phase of the two points is the same since y is the same
kx1-7t1 = kx2- 7t2
7(t2 -t1) = k(x2 -x1)
If t2>t1 then x2>x1
As t <0, v =x/ t < 7/k
Summary
Chapter 17 Problem 12
• 7 = 2%f = 2%/T
k= 2%/
• v = 7/k = f = /T
• wave speed= one wavelength per period
• y(x,t)=ym sin( kx- 7t) describes a wave
moving right at constant speed v= 7/k
• kx- 7t = const labels a point on the wave
x = (7/k) t + const
• y(x,t)=ym sin( kx+ 7t) is a wave moving left
• kx+ 7t = const labels a point on the wave
x = -(7/k) t + const
• The equation of a transverse wave travelling
on a string is
• y(x,t) = 6.0 sin (0.020 %x + 4.0 %t) where
x is in cm, y is in cm and t in seconds
• (a) amplitude?
• ym = 6.0 cm
• (b) wavelength?
• k=2%/ = .02 % = .01 (2%) hence =100 cm
• (c) frequency?
• 7= 4.0 % = 2%f hence f= 2 Hz
Problem 12 Cont’d
y(x,t) =6.0sin(0.020 %x + 4.0 %t)
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speed?
v= f = /T = (100cm)(2 Hz) = 200 cm/s
direction?
left
maximum transverse speed of a particle in
the string?
• vy(x,t)= dy/dt = 24.0%cos(0.020%x +4.0 %t)
vymax = 24.0 % cm/s
Wave speed of a stretched string
• Actual value of v = 7/k is determined by the
medium
• as wave passes, the “particles” in the medium
oscillate
• medium has both inertia (KE) and elasticity (PE)
• dimensional argument: v= length/time LT-1
• inertia is the mass of an element =mass/length ML-1
• tension - is the elastic character (a force) MLT-2
• how can we combine tension and mass density to get
units of speed?
2
Wave speed of a stretched string
• v = C (-/)1/2
(MLT-2/ML-1)1/2 =L/T
• detailed calculation using 2nd law yields C=1
v = (-/)1/2
• speed depends only on characteristics of string
• independent of the frequency of the wave
- f due to source that produced it
• once f is determined by the generator, then
•
= v/f = vT
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