Comparing rate of change by period: A sine wave with period 4 ms changes faster than a sine wave with which of the following periods?

Introduction / Context:
Since frequency f = 1/T, longer period means lower frequency and thus a slower rate of change for a sinusoid of the same amplitude. This question requires translating between period values and qualitative 'faster/slower' behavior—an everyday skill in timing and signal design.

Given Data / Assumptions:

• Reference period: T_ref = 4 ms.
• Candidate periods: 0.0045 s (4.5 ms), 2 ms, 1.5 ms, 3000 µs (3 ms).
• Equal amplitudes assumed for fair slope comparison.

Concept / Approach:
Shorter period → higher frequency → faster change. Therefore, a sinusoid with T_ref = 4 ms will change faster than any sinusoid with a longer period (T > 4 ms) and slower than those with shorter periods (T < 4 ms).

Step-by-Step Solution:

Verification / Alternative check:
Compute frequencies: f(4 ms) = 250 Hz; f(4.5 ms) ≈ 222 Hz. Since 250 > 222, the 4 ms waveform has higher dv/dt for the same amplitude.

Why Other Options Are Wrong:

• 2 ms, 1.5 ms, 3000 µs (3 ms): All are shorter periods, hence faster than 4 ms; they do not satisfy '4 ms is faster than them'.

Common Pitfalls:

• Mistakes converting µs to ms and s, causing reversed comparisons.

Final Answer:
0.0045 s