standing waves on a string

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Standing waves on a string Key concepts: a string that is plucked with both ends fixed results in standing waves on the string (aka normal modes) o amplitudes are 0 at x=0 and x=L (L = length) fundamental frequency aka first harmonic: the lowest frequency corresponding to the longest wavelength (λ = 2L) wavelength: ! = 2/ where m is a positive, nonzero integer fundamental frequency: ! = ! ! = ! !! = ! !! ! ! , where v = wavespeed, T = tension, = linear mass density frequency: fm =mf1 Example: A guitar string is 0.61m long, has a fundamental frequency of 500Hz, and a tension kept at 80.0N a) Find the wave speed of the string (hint: find linear mass density) b) Find the wavelengths and frequencies for the 2 nd ,3 rd , and 4 th normal modes of vibration ANSWER a) = ! ! ! = ! !! ! ! → 500 = ! !(!.!"!) ∗ !".! ! = 0.0013kg/m = = 80.0 0.013 = 245m/s b) for the 2 nd harmonic: wavelength: ! = 2/ ! = !! ! = !(!.!") ! = 0.61 frequency: fm =mf1 ! = 2 500 = 1000Hz For the 3 rd harmonic: wavelength: ! = 2/ ! = !! ! = !(!.!") ! = 0.41 frequency: fm =mf1 ! = 3 500 = 1500Hz For the 4 th harmonic: wavelength: ! = 2/ ! = !! ! = !(!.!") ! = 0.31 frequency: fm =mf1 ! = 4 500 = 2000Hz

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Page 1: Standing waves on a string

Standing waves on a string Key concepts:

-‐ a string that is plucked with both ends fixed results in standing waves on the string (aka normal modes)

o amplitudes are 0 at x=0 and x=L (L = length) -‐ fundamental frequency aka first harmonic: the lowest frequency

corresponding to the longest wavelength (λ = 2L) -‐ wavelength: 𝜆! = 2𝐿/𝑚 where m is a positive, nonzero integer

-‐ fundamental frequency: 𝑓! = !!= !

!!𝑣 = !

!!!!, where v = wavespeed, T =

tension, 𝜇 = linear mass density -‐ frequency: fm = mf1

Example: A guitar string is 0.61m long, has a fundamental frequency of 500Hz, and a tension kept at 80.0N

a) Find the wave speed of the string (hint: find linear mass density) b) Find the wavelengths and frequencies for the 2nd, 3rd, and 4th normal modes

of vibration ANSWER

a) 𝑣 = !!

𝑓! = !!!

!!→ 500 = !

!(!.!"!)∗ !".!

!

𝜇 = 0.0013kg/m

𝑣 = 𝑇𝜇 =

80.00.013 = 245m/s

b) for the 2nd harmonic:

wavelength: 𝜆! = 2𝐿/𝑚 à 𝜆! =!!!= !(!.!")

!= 0.61𝑚

frequency: fm = mf1 à 𝑓! = 2 500 = 1000Hz For the 3rd harmonic: wavelength: 𝜆! = 2𝐿/𝑚 à 𝜆! =

!!!= !(!.!")

!= 0.41𝑚

frequency: fm = mf1 à 𝑓! = 3 500 = 1500Hz For the 4th harmonic: wavelength: 𝜆! = 2𝐿/𝑚 à 𝜆! =

!!!= !(!.!")

!= 0.31𝑚

frequency: fm = mf1 à 𝑓! = 4 500 = 2000Hz

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