Assuming signal A has a lower frequency than signal B, which statement is true?

Assuming signal A has a lower frequency than signal B, which statement is true?

• A. A has a longer period than B.
• B. A has a lower frequency than B.
• C. A has a shorter wavelength than B.
• D. A has a better axial resolution than B.

Answer: (A)

The main idea is that period and frequency are inverses of each other: T = 1/f. If signal A has a lower frequency than signal B, then its period must be longer (since a slower oscillation takes more time to complete a cycle). So A has a longer period than B. If we consider wavelength in the same medium, wavelength is λ = v/f, so lower frequency means a longer wavelength, not shorter. And axial resolution typically improves with higher frequency (shorter wavelengths), so lower frequency would not give better resolution. With those points in mind, the statement that A has a longer period than B is the correct one.

The main idea is that period and frequency are inverses of each other: T = 1/f. If signal A has a lower frequency than signal B, then its period must be longer (since a slower oscillation takes more time to complete a cycle). So A has a longer period than B.

If we consider wavelength in the same medium, wavelength is λ = v/f, so lower frequency means a longer wavelength, not shorter. And axial resolution typically improves with higher frequency (shorter wavelengths), so lower frequency would not give better resolution. With those points in mind, the statement that A has a longer period than B is the correct one.