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If the base period is 2π, what frequency does cos⁡(2t)\cos (2t) have?

If the base period is 2π2\pi, the term cos⁡(t)\cos(t) has a frequency of 12π\frac{1}{2\pi} cycles per second. Doubling the argument to cos⁡(2t)\cos(2t) doubles the frequency, so cos⁡(2t)\cos(2t) has a frequency of 1π\frac{1}{\pi} cycles per second.

Conditions

  • The base period of the function is 2π2\pi.
  • The frequency is defined as the reciprocal of the period.

Reasoning, step by step

  1. Identify the base frequency of cos⁡(t)\cos(t) as 12π\frac{1}{2\pi}.
  2. Recognize that the argument 2t2t implies a period of π\pi.
  3. Calculate the frequency of cos⁡(2t)\cos(2t) as 1π\frac{1}{\pi}.
  4. Conclude that doubling the argument doubles the frequency.

Example

The speaker says, 'This has a frequency of 1 over 2 pi. This has twice the frequency. This has a frequency of 1 over pi.'

Common misconceptions

  • Treating cos⁡(2t)\cos(2t) as just another cosine without recognizing that the 2 changes the frequency.
  • Assuming the frequency remains 12π\frac{1}{2\pi} for all harmonic terms.

Watch the explanation

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