A string of length 0.5 m is held fixed at both ends. If the mass per unit length of the string is 1 g m-1 and the tension in the string is 50 N. What is the frequency (in Hz) of the 5th standing wave mode of the string?

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BlackTom AI · Accepted Answer

The problem describes a string of length L = 0.5 m, fixed at both ends, with linear mass density μ = 1 g/m = 0.001 kg/m, and tension T = 50 N. For a string fixed at both ends, the standing-wave frequencies are given by f_n = n/(2L) * sqrt(T/μ), where n is the mode number.

First, compute the wave speed component sqrt(T/μ): sqrt(T/μ) = sqrt(50 / 0.001) = sqrt(50000) ≈ 223.6068 m/s.

Next, evaluate the factor 1/(2L): 2L = 2 * 0.5 m = 1.0 m, so 1/(2L) = 1.0.

Therefore, f_n = n * 1/(2L) * sqrt(T/μ) = n * 223.6068 Hz.

Now consider each option:

Option a: 223.6 Hz corresponds to n = 1 (the fundamental) since f_1 ≈ 1 * 223.6068 Hz. This would be the first mode, not the 5th, so it describes a lower frequency than asked.

Option b: 1118.0 Hz matches the 5th mode, since f_5 ≈ 5 * 223.6068 ≈ 1118.034 Hz. This is consistent with the formula for n = 5 and the given parameters.

Option c: 447.2 Hz would correspond to n ≈ 2, because 2 * 223.6068 ≈ 447.2136 Hz. This is the second harmonic, not the fifth.

To summarize, the key steps are identifying the correct formula for a string with fixed ends, computing the wave speed sqrt(T/μ), applying f_n = n/(2L) * sqrt(T/μ), and then matching the value to the given mode number. The calculated 5th mode frequency is approximately 1118 Hz, which aligns with option b.

PHAS0005_24-25 Quiz - week 3

A string of length 0.5 m is held fixed at both ends. If the mass per unit length of the string is 1 g m-1 and the tension in the string is 50 N. What is the frequency (in Hz) of the 5th standing wave mode of the string?

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