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STANDING WAVES ON A
STRING
Jasmine Keller
27756148
LF2 Green
The Basics
Where L is the length of the string Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics-
of-Standing-Waves
Must be in slideshow format to view gif
Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics-of-Standing-Waves
Normal Modes
• Standing waves called normal modes of the vibration of a
string have frequencies corresponding to:
• fm = m/2L *√(T/μ)
• The fundamental frequency is proportional to the square
root of the tension in the string and inversely proportional
to its length and to the square root of the linear mass
density.
• Higher frequencies correspond to higher values of m:
• fm = mf1 m = 1,2,3,4….
• Therefore the first few allowed harmonics are:
• f1 = 1/2L √(T/μ)  f2 = 2f1  f3 = 3f1 and so on…
Test Your Knowledge
• The string below is 1.5 meters long and is vibrating as the
first harmonic. The string vibrates up and down with 33
complete vibrational cycles in 10 seconds. Determine: the
frequency, period, wavelength and speed for this wave.
Frequency  f =
Period  T =
Wavelength  λ =
Speed  v =
Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics-
of-Standing-Waves
Answer
• Frequency  f =3.3 Hz
• Period  T = 0.303 seconds
• Wavelength  λ = 3.0 m
• Speed  v =9.9m/s
Given: L = 1.5 m
33 cycles in 10 seconds
The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of
cycles per unit of time. In this case, it is
f = (33 cycles) / (10 seconds) = 3.3 Hz
The period is the reciprocal of the frequency.
T = 1 / (3.3 Hz) = 0.303 seconds
The wavelength of the wave is related to the length of the rope. For the first harmonic as pictured in this problem, the length of the
rope is equivalent to one-half of a wavelength. That is, L = 0.5 • W where W is the wavelength. Rearranging the equation and
substituting leads to the following results:
W = 2 • L = 2 • (1.5 m) = 3.0 m
The speed of a wave can be calculated from its wavelength and frequency using the wave equation:
v = f • W = (3.3 Hz) • (3. 0 m) = 9.9 m/s

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LO6- Standing waves on a string

  • 1. STANDING WAVES ON A STRING Jasmine Keller 27756148 LF2 Green
  • 2. The Basics Where L is the length of the string Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics- of-Standing-Waves
  • 3. Must be in slideshow format to view gif Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics-of-Standing-Waves
  • 4. Normal Modes • Standing waves called normal modes of the vibration of a string have frequencies corresponding to: • fm = m/2L *√(T/μ) • The fundamental frequency is proportional to the square root of the tension in the string and inversely proportional to its length and to the square root of the linear mass density. • Higher frequencies correspond to higher values of m: • fm = mf1 m = 1,2,3,4…. • Therefore the first few allowed harmonics are: • f1 = 1/2L √(T/μ)  f2 = 2f1  f3 = 3f1 and so on…
  • 5. Test Your Knowledge • The string below is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine: the frequency, period, wavelength and speed for this wave. Frequency  f = Period  T = Wavelength  λ = Speed  v = Source: http://www.physicsclassroom.com/class/waves/Lesson-4/Mathematics- of-Standing-Waves
  • 6. Answer • Frequency  f =3.3 Hz • Period  T = 0.303 seconds • Wavelength  λ = 3.0 m • Speed  v =9.9m/s Given: L = 1.5 m 33 cycles in 10 seconds The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. In this case, it is f = (33 cycles) / (10 seconds) = 3.3 Hz The period is the reciprocal of the frequency. T = 1 / (3.3 Hz) = 0.303 seconds The wavelength of the wave is related to the length of the rope. For the first harmonic as pictured in this problem, the length of the rope is equivalent to one-half of a wavelength. That is, L = 0.5 • W where W is the wavelength. Rearranging the equation and substituting leads to the following results: W = 2 • L = 2 • (1.5 m) = 3.0 m The speed of a wave can be calculated from its wavelength and frequency using the wave equation: v = f • W = (3.3 Hz) • (3. 0 m) = 9.9 m/s