Standing waves can be described by the equation D(x,t) = 2A sin(kx) cos (ωt). Nodes occur where the amplitude is zero and antinodes where it is maximum. On a vibrating string, the wavelength is determined by the length of the string and tension. The fundamental frequency produces the longest wavelength with no nodes. Higher harmonics have shorter wavelengths and more nodes. Examples calculate the frequency of a guitar string at different tensions and lengths of the standing wave where the amplitude is a given value.
2. StandingWaves
A standing wave is described by the equation
D(x,t) = 2A sin(kx) cos (ωt)
D(x,t) = 2A sin(2πx/λ) cos (2πt/T)
The amplitude A(x) depends on the position of the standing
wave
A(x) = 2A sin(kx) = 2A sin(2πx/λ)
Combining the two equations creates:
D(x,t)= A(x)cos(ωt)
3. Amplitude – Nodes and Antinodes
Nodes: points where the amplitude is zero
A(x) = 0
Consecutive nodes are half a wavelength apart
Antinodes: point of maximum possible amplitude of 2A
A(x)=2A
Consecutive antinodes are half a wavelength apart
Adjacent antinodes and nodes are a quarter of a wavelength
apart
4. StandingWaves on Strings
Two ends of a string, with length L, are fixed at both ends.
The amplitude at each end must be zero (when x=0 and when
x=L)
A(x)= 2A sin(2πx/λ) = 2A sin(2πL/λ) = 0
Since a node repeats ever half a wavelength or π
2πL/λ = mπ Where m is a positive nonzero integer
Rearranging the equation can help find frequency:
f=v/λ = mv/2L= m/2L√(T/μ)
5. The lowest frequency(f1) produces the longest wavelength, where
λ=2L and is the first harmonic. It has zero nodes. It is also called
fundamental frequency or first harmonic
Each consequent frequency is given by the equation
fm=mf1
Where ‘m’ refers to the positive nonzero integers
‘m’ also indicates the subsequent harmonics or resonant
frequencies. f2 =2f1 is the second harmonic.
6. Question Set 1
The D string on a guitar has a length of 0.70m and a linear
mass density of 7.86 Χ 10-4kg/m.To maintain the D note on
the guitar, the string is kept at a tension of 98N.
A)What is the frequency when the wave length is half it’s
original length
B)What is the frequency if the tension is decreased by 10N.
7. Answer Set 1
A) Using the equation: f= m/2L√(T/μ).The length is half the original
so L equals to 0.35m.TheTensions is given as 98N and the linear
mass density is 7.86 Χ 10-4kg/m.
The frequency becomes roughly 504Hz
B)The tension is decreased by 10N to result in a new tension of
88N. Similar to the above question, the equation: f= m/2L√(T/μ) is
used, but this time the tension is 88N rather than 98N.
The frequency is roughly 239Hz.
8. Question Set 2
A standing wave has a wavelength of 25cm and a frequency
of 9Hz. It has an antinode that reaches up to 4cm.
A) Determine the standing wave equation of the form:
D(x,t) = 2A sin(kx) cos (ωt)
B) Determine the first position where the amplitude first
reaches 1.6cm
9. Answer Set 2 – Pt.1
A)The ‘k’ and the ‘ω’ in the equation for D(x,t) can be found
by:
K= 2π/λ, ω=2π/T and whereT=1/f
k= 2π/25 ≈.025
T= 1/9, ω=2π/T = 2π(9) ≈ 56.55
The antinode corresponds to the ‘2A’ of the equation D(x,t) and it
equals to 4cm
Plugging these values into the equation gives the answer:
D(x,t) = 4 sin(.25x) cos (56.55t)
10. Answer Set 2 – Pt.2
B) Using the equation: A(x) = 2A sin(kx) = 2A sin(2πx/λ), and the
results from above we can find the position (x) where the amplitude
is first at 1.6cm.
1.6cm = 4sin(.25x)
X≈ 1.64cm
