For the nth harmonic with two fixed ends, lambda_n = 2L/n. What is lambda_1?

For the nth harmonic with two fixed ends, lambda_n = 2L/n. What is lambda_1?

• A. L
• B. L/2
• C. 4L
• D. 2L

Answer: (D)

Two fixed ends impose nodes at both ends, so the standing wave must contain an integer number of half-wavelengths along the length. For the fundamental (the first harmonic) there is just one half-wavelength in the length, so L equals lambda_1 divided by 2. Solve for lambda_1: lambda_1 = 2L. This matches the given formula when n = 1. The other numbers would correspond to higher harmonics, not the first, so the correct wavelength is 2L.

Two fixed ends impose nodes at both ends, so the standing wave must contain an integer number of half-wavelengths along the length. For the fundamental (the first harmonic) there is just one half-wavelength in the length, so L equals lambda_1 divided by 2. Solve for lambda_1: lambda_1 = 2L. This matches the given formula when n = 1. The other numbers would correspond to higher harmonics, not the first, so the correct wavelength is 2L.