A powerpoint presentation about standing waves Amplitude, what a standing wave, what anti-nodes and nodes are and what the amplitude of a standing wave is. Specific examples with calculations are provided.

1. STANDING WAVES
Sections 14.2 & 14.3 of the text
By: Harkirat Dhaliwal

2. WHAT ARE STANDING
WAVES?
A standing wave is a vibration of a system in which some
particular points remain fixed (such as frequency), while
other points vibrate with a maximum amplitude.

3. WHAT ARE NODES AND
ANTINODES?
Points of no displacement in a standing wave pattern are
referred to as nodes.
Points of maximum displacement are known as antinodes.
Exact location of nodes of a standing wave can be found
using the equation: x= m(wavelength/2) where m is 0, +-1,
+-2……..
Exact location of antinodes can be found using the equation
x = (m+0.5)(wavelength/2) where m is 0, +-1, +-2………

4. QUESTION 1
On the diagram above, which of the labels represents an
antinode?
A
B
C
A
B
C

5. QUESTION 1
On the diagram above, which of the labels represents an
antinode?
A -is the correct answer as it represents the
maximum displacement of the standing
wave
A
B
C

6. AMPLITUDE OF STANDING
WAVE
Amplitude of a standing wave is position dependent.
It is represented by the formula :
A(x) = 2Asin(kx) = 2A sin(2pi(x)/wavelength)

7. QUESTION 2
The amplitude of a standing wave is given by:
A(x) = (32.0cm)sin(4.00x)
a) Determine the amplitude and wavelength of the travelling
waves.
b) Determine the location of the first 2 nodes and first 2
antinodes along the positive x-axis.
c) Determine the first point from the origin where the
amplitude is 0.20m.

8. QUESTION 2-SOLUTION
A(x) = 2Asin(kx) = 2A sin(2pi(x)/wavelength)
A(x) = (32.0cm)sin(4.00x)
a) So according to equation above: 2A= 32.0 cm, so A= 32.0/2= 16.0 cm (amplitude)
Wavelength: 2pi/wavelength= 4.00 rad/m
2pi= 4 x wavelength
2pi/4 = wavelength
pi/2 m= wavelength

9. QUESTION 2- SOLUTION
CONTINUED
b) First 2 nodes occur when sin(2pi(x)/wavelength)= 0.
2pix/wavelength= mpi (where m= 0, 1, 2..)
x= m (wavelength/2)= m((pi/2)/2)
when m= 0, x=0m
when m=1, x= pi/2 x (1/2)= pi/4 m
So first 2 nodes occur at 0 m & 0.785 m.

10. QUESTION 2-SOLUTION
CONTINUED
c) Set amplitude= 0.20 m and solve for x
0.20 m= (0.32m) sin(4.00x)
0.20/0.32= sin(4.00x)
0.625m= sin(4.00x)
sin^-1(0.625) (sin inverse) = 4.00x
0.675 rad= 4x
0.675/4= x= 0.17 m

11. THANK YOU!