16.2 Period and frequency in oscillations

Determine the frequency of two oscillations: medical ultrasound and the period of middle c

We can use the formulas presented in this module to determine both the frequency based on known oscillations and the oscillation based on a known frequency. Let’s try one example of each. (a) A medical imaging device produces ultrasound by oscillating with a period of 0.400 µs. What is the frequency of this oscillation? (b) The frequency of middle C on a typical musical instrument is 264 Hz. What is the time for one complete oscillation?

Strategy

Both questions (a) and (b) can be answered using the relationship between period and frequency. In question (a), the period $T$ is given and we are asked to find frequency $f$ . In question (b), the frequency $f$ is given and we are asked to find the period $T$ .

Solution a

1. Substitute $0\text{.}\text{400}\mathrm{\text{μ}}\text{s}$ for $T$ in $f=\frac{1}{T}$ :
   $f=\frac{1}{T}=\frac{1}{0\text{.}\text{400}\times {\text{10}}^{-6}\mathrm{s}}.$

   Solve to find

   $f=2\text{.}\text{50}\times {\text{10}}^6\text{Hz}.$

Discussion a

The frequency of sound found in (a) is much higher than the highest frequency that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb.

Solution b

1. Identify the known values:

   The time for one complete oscillation is the period $T$ :

   $f=\frac{1}{T}.$
2. Solve for $T$ :
   $T=\frac{1}{f}.$
3. Substitute the given value for the frequency into the resulting expression:
   $T=\frac{1}{f}=\frac{1}{\text{264}\text{Hz}}=\frac{1}{\text{264}\text{cycles/s}}=3\text{.}\text{79}\times {\text{10}}^{-3}\text{s}=3\text{.}\text{79}\text{ms}.$

Discussion

The period found in (b) is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case).