College

Two strings are respectively 1.00 m and 2.00 m long. Which of the following wavelengths, in meters, could represent harmonics present on both strings?

1) 0.800, 0.670, 0.500
2) 1.33, 1.00, 0.500
3) 2.00, 1.00, 0.500
4) 2.00, 1.33, 1.00
5) 4.00, 2.00, 1.00

Answer :

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

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).