Standing Wave Presentation

1. Standing Waves
Rachel Sunderland

2.  Two waves that are moving in opposite directions
with the same amplitude, wavelength, and frequency
 They result from the superstition of waves when a
wave reflects from a surface and interferes with the
wave travelling in the original direction
What is a Standing Wave?

3.  D1(x,t) = A sin(kx + ωt) <- occurs at increasing x
 D2(x,t) = A sin(kx - ωt) <- occurs at decreasing x
 Together, the equation of a standing wave becomes:
 D(x,t) = D1(x,t) + D2(x,t)
 = A sin(kx – ωt) + A sin(kx + ωt)
 = A sin(kx – wt) + sin(kx – wt)
 = 2A sin(kx) cos (ωt)
Standing Wave Equations

4.  In this photo, the wave on the top is moving to the
right and the wave to the bottom is moving to the left
 They combine together in order to form one wave
with an amplitude of 2A at the anti-node and 0 at the
node
Standing Waves Visual Example

5.  A node is the point where the amplitude and thus,
there is no displacement from equilibrium
 The antinode is the point where there is the
maximum displacement from equilibrium and thus,
the amplitude is also maximum
 Since a standing wave is two waves moving in
opposite directions, the maximum amplitude of a
standing wave is twice the amplitude of one wave
What is a Node?

6. Picture of Nodes/Anti-nodes
 A node is where the displacement is zero and an anti-
node is where the displacement is 2A as seen in the
photograph below

7.  Nodes occur when the displacement is zero
 Therefore, sin (2πx/𝜆) = 0
 Since is zero at every pi interval so we can substitute
it in for 0
 2πx/𝜆 = mπ where m = positive or negative (1,2,3…)
 In order to simplify, we bring x to one side, and cancel
out the π’s in order to get the equation:
 X = m𝜆/2
Calculating the Position of a Node

8.  At what position does the wave function:
0.75m sin(17.2x) have its first node?
 We use the equation: X = m𝜆/2
 1st we need to solve for the wavelength
 We know that k = 2π/𝜆
 Therefore, 17.2 = 2π/𝜆 and 𝜆 = 0.365
 We can then plug it into the equation x = 1(0.365/2) =
0.1825
Example

9.  Anti-Nodes occur when the displacement is (+/-) 2A
 Therefore, sin (2πx/𝜆) = (+/-) 1
 Since (+/-) occurs at every pi/2 interval so we can
substitute it in for (+/-) 1
 2πx/𝜆 = (m+ 1/2)π where m = positive or negative
(1,2,3…) which also equals (m+1/2) *𝜆/2
 In order to simplify, we bring x to one side in order to
get the equation:
 X = (+/-) 𝜆/4, 3𝜆/4, 5𝜆/4 …
Calculating the Position of an
Anti-Node

10.  At what position does the wave function:
0.50m sin(5.9x) have its first anti-node?
 We use the equation: X = (+/-) 𝜆/4, 3𝜆/4
 1st we need to solve for the wavelength
 We know that k = 2π/𝜆
 Therefore, 5.9 = 2π/𝜆 and 𝜆 = 1.07
 We can then plug it into the equation x = 1(1.07/2) =
0.535
Example

11.  What is the amplitude of a wave when its at the
location x = 0.35m, and its wavelength is 2.13m
 1st we need to determine what the value of K is
 We can use the equation K = 2π/𝜆
 Therefore, k = 2π/2.13 = 2.95 rad/m
 We can then use this information to complete the
formula
Integrated Example

12.  A(x) = 2A sin (2π/𝜆)
 Plugging in values gives us
 0.35m = 2(x) sin (2.95)
 0.35m = 2x (0.0514)
 6.81 = 2x
 x = 3.41
 Therefore, at the location x = 0.35m and with a wavelength
of 2.13m, the amplitude of the wave is 3.41m
Integrated Example Part 2

13.  "S-cool, the Revision Website." Standing Waves. N.p.,
n.d. Web. 09 Mar. 2015.
 "Standing Waves." Mr Magarey's Official Year 11 Physics
Class Wikispace -. NHS, n.d. Web. 09 Mar. 2015.
Sources Cited