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Answer :
Final answer:
Option (3), with wavelengths of 2.00 m, 1.00 m, and 0.500 m, represents the harmonics that could be present on both a 1.00 m and a 2.00 m long string, as they fit as multiples of the string lengths.
Explanation:
When dealing with harmonics on strings, the wavelengths of standing waves that can exist on the string are directly related to the length of the string. For a string that is 1.00 m long, wavelengths that are multiples of 2 meters (2, 1, 0.666..., etc.) can form standing waves. Similarly, for a 2.00 m long string, these will be multiples of 4 meters. We can find common harmonics that can exist on both strings by identifying wavelengths that are multiples of both 2 and 4 meters.
Looking at the options provided, only set (3) which consists of wavelengths 2.00 m, 1.00 m, and 0.500 m matches our requirement. 2.00 m and 1.00 m are the first and second harmonics of the 1.00 m string (n = 1 and n = 2) and also the first harmonic of the 2.00 m string (n = 1). The 0.500 m wavelength is the fourth harmonic for the 1.00 m string (n = 4) and second harmonic for the 2.00 m string (n = 2).
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Answer:
5) 4.00, 2.00, 1.0
Explanation:
wave equation is given as;
F₀ = V / λ
Where;
F₀ is the fundamental frequency = first harmonic
Length of the string for first harmonic is given as;
L₀ = (¹/₂) λ
λ = 2 L₀
when L₀ = 1
λ = 2 x 1 = 2m
when L₀ = 2m
λ = 2 x 2 = 4m
For First harmonic, the wavelength is 2m, 4m
For second harmonic;
L₁ = (²/₂)λ
L₁ = λ
When L₁ = 1
λ = 1 m
when L₁ = 2
λ = 2 m
For second harmonic, the wavelength is 1m, 2m
Thus, the wavelength that could represent harmonics present on both strings is 4m, 2m, 1 m
