If the length L is halved on a string fixed at both ends with constant wave speed, what happens to the fundamental frequency f1 = v/(2L)?

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Multiple Choice

If the length L is halved on a string fixed at both ends with constant wave speed, what happens to the fundamental frequency f1 = v/(2L)?

Explanation:
The key idea is that the fundamental frequency of a string fixed at both ends scales inversely with its length when the wave speed is fixed. The fundamental mode fits one half-wavelength into the string, so λ1 = 2L and f1 = v/λ1 = v/(2L). If the length is halved, λ1 becomes 2(L/2) = L, and the frequency becomes f1' = v/λ1' = v/L, which is twice the original f1. So, halving the length doubles the fundamental frequency.

The key idea is that the fundamental frequency of a string fixed at both ends scales inversely with its length when the wave speed is fixed. The fundamental mode fits one half-wavelength into the string, so λ1 = 2L and f1 = v/λ1 = v/(2L). If the length is halved, λ1 becomes 2(L/2) = L, and the frequency becomes f1' = v/λ1' = v/L, which is twice the original f1. So, halving the length doubles the fundamental frequency.

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