L E A R N I N G O B J E C T V IJ E N N Y H E

STANDING WAVES

WHAT IS A STANDING WAVE?

Waves with equal amplitude, wavelength, and frequency travelling in opposite directions in the same medium

http://langlopress.net/homeeducation/resources/science/content/support/illustrations/Standing%20Wave/

Wave moving right Wave moving left D1 = Asin(kx-wt) D2 = Asin(kw+wt)

SUPERPOSITION (D 1 + D 2 )

STANDING WAVE*D(x,t) = 2Asin(kx)cos(wt)

D(x,t) = 2Asincos(2πft)

***By the Principle of superposition, the resultant wave is the sum of the individual waves: D1 + D 2

EQUATION

*Simplified using Trig. Identity:sin(a-b)+sin(a+b)=2sin(a)cos(b)

EQUATION APPLICATION

• According to the standing wave equation, there is a position dependent amplitude:

A(x) = 2Asin(kx) = 2Asin• So, we can simplify the standing wave equation to:

D(x,t) = A(x)cos(wt)

• When D(x,t) = 0 , there is zero amplitude and therefore exists a NODE

• Similarly, when D(x,t) = 2A, this denotes maximum amplitude and therefore exists an ANTINODE

WHERE ARE THEY LOCATED?

NODES

Occur at A(X)= 2Asin= 0Solving for x, we get

x = m, where m=0, ±1, ±2…Simplify to get: x=m ,

m=0, ±1, ±2…So, nodes are found at

x= 0, ±, ± ±…

ANTINODES

Occur at A(X)= 2Asin=2A Solving for x, we get

x = (m+ ), m=0, ±1, ±2…Simplify to get: x=(m+)

m=0, ±1, ±2…So, antinodes are found

at X= ±, ± , ±…

RECAP

NODE - ANTINODE -

2A

V

Vhttp://www.met.reading.ac.uk/pplato2/h-flap/phys5_6.html

STANDING WAVES ON STRINGS

• When a string is attached to two fixed ends, waves will reflect from the fixed ends and travel in opposite directions: STANDING WAVE!

• If one end is denoted as the origin and the other end is denoted x=L, the amplitude at the origin and at x=L is 0.

sin= 0 substitute x=L

m = 1, 2, 3, 4

A string with fixed ends will oscillate with wavelength:

m = 1, 2, 3, 4

FUNDAMENTAL FREQUENCY

• Frequency is related to velocity and wave length by: f= ; substitute in

f= ; V is related to tension and mass density by: √()

So, f= √()• When the longest wavelength (2L) and lowest

frequency are achieved, it is defined as the fundamental frequency (first harmonic)

f = √()

HARMONICS

• These standing waves are referred as normal modes of vibration of a string

• Higher frequencies are often multiples of fundamental frequency: fm = mf1 m=1, 2, 3,4…

• These frequencies are known as harmonics or resonant frequencies

Fundamental harmonic: f1

Second harmonic: 2f1

Third harmonic: 3f1

Fourth harmonic: 4f1

= 2L

= L

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html

PRACTICE PROBLEMS

Q1: A 65.0 cm long, 7.0 g string harp string is strummed by a harpist. It oscillates in its m=4 mode of vibration with a frequency of 200 Hz and an amplitude of 5 cm. Determine a) wavelength of the wave and b) tension in the string.

http://galleryhip.com/harpist-clipart.html

SOLUTION TO Q1

• First of all, we know that the mode is m=4. Which equation will allow us to determine the wavelength? =

• We are given that the length of string is 65.0 cm So = = 32.5 cm = 0.325

m

• Now we are asked to determine the tension of the string. Which equations will be useful to use? 1. V= √() and 2. v= f

• By rearranging the 1st equation, we get T= V2 . Do we know what v and are?

• Since we are given the mass and length of string, we can first determine • = = 0.0108 kg/m

• Now we are only missing V. What other information are we given? We are given the frequency, and we calculated the wavelength in a. Therefore, we can use v= f relationship

SOLUTION TO Q1 (CONT.)

• v= f = (0.325 m) (200 Hz) = 65.0 m/s

• Now we have all the required information to solve for T= V2

T = (65.0 m/s) 2 (0.0108 kg/m) T = 45.6 N

PRACTICE PROBLEMS

Q2: A ukelele string oscillates in a standing wave pattern by :

D(x,t) = 2.0 cm (sin(0.5x))(cos(10π)t)X= meters t=seconds

a) What is the wavelength and frequency of the standing wave?

b) Determine the distance between two consectuive nodes and between a node and an adjacent antinode.

c) What is the displacement for a particle located at x=0.1m when t=3 s?

SOLUTION TO Q2

a) To determine the wavelength, we can use the relationship: k=From the given equation, we know k=0.5 rad/mSo, = = 4πm

Next, we know the value of w from the equation, so we can use w= 2 f to solve for frequency. f= = = 5.0 Hz

b) The distance between two consecutive nodes is half a wavelength. Knowing this, we can divide the wavelength calculated in part a by two= 2π m

The distance between two consecutive antinodes is quarter of a wavelength. Therefore, we can multiply ¼ to 4 and we get m.

SOLUTION TO Q2 (CONT.)

c) The given equations is:

D(x,t) = 2.0 cm (sin(0.5x))(cos(10π)t)

To evaluate the displacement at x=0.1m and t=3.0 s, we simply plug in the values into the given equation.

D(0.1m, 3.0s) = 0.02 m[sin(0.5(0.1m))][cos(10π(3s))]

= 0.00099958… = 0.001 m or 1 mm