Period and Frequency in Oscillations

The given figure shows a closed zoom view of the strings of a guitar. There are two slanting white colored strings in the picture. In the nearer string, the gaps between the circular threads of the string are visible, whereas the second white string at the back looks like a white thin stick.

When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each successive vibration of the string takes the same time as the previous one. We define periodic motion to be a motion that repeats itself at regular time intervals, such as exhibited by the guitar string or by an object on a spring moving up and down. The time to complete one oscillation remains constant and is called the period $T$ . Its units are usually seconds, but may be any convenient unit of time. The word period refers to the time for some event whether repetitive or not; but we shall be primarily interested in periodic motion, which is by definition repetitive. A concept closely related to period is the frequency of an event. For example, if you get a paycheck twice a month, the frequency of payment is two per month and the period between checks is half a month. Frequency $f$ is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is

$$f=\frac{1}{T}. $$

The SI unit for frequency is the cycle per second, which is defined to be a _ hertz_ (Hz):

$$1 \text{Hz}=1\frac{\text{cycle}}{\text{sec}}\text{ or } 1 \text{Hz}=\frac{1}{\text{s}} $$

A cycle is one complete oscillation. Note that a vibration can be a single or multiple event, whereas oscillations are usually repetitive for a significant number of cycles.

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 $0.400 \mathrm{\text{μ}}\text{s}$ for $T$ in $f=\frac{1}{T}$ :
    $$f=\frac{1}{T}=\frac{1}{ 0.400 \times 10^{-6} s}. $$

    Solve to find

    $$f= 2.50 \times 10^{6} \text{Hz}. $$

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:
    $$T=\frac{1}{f}=\frac{1}{264 \text{Hz}}=\frac{1}{264 \text{cycles/s}}= 3.79 \times 10^{-3} \text{s}= 3.79 \text{ms}. $$

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).

Check your Understanding

Identify an event in your life (such as receiving a paycheck) that occurs regularly. Identify both the period and frequency of this event.

I visit my parents for dinner every other Sunday. The frequency of my visits is 26 per calendar year. The period is two weeks.

Section Summary

Problems & Exercises

What is the period of $60.0 \text{Hz}$ electrical power?

Strategy

Period and frequency are reciprocals: $T = \frac{1}{f}$.

Solution

$$T = \frac{1}{f} = \frac{1}{60.0 \text{ Hz}} = 0.0167 \text{ s} = 16.7 \text{ ms}$$

Discussion

The 60 Hz frequency is standard for electrical power in North America. This means the voltage oscillates with a period of 16.7 ms, completing 60 full cycles every second. In countries using 50 Hz power, the period would be 20 ms. This rapid oscillation is why incandescent lights flicker at twice this frequency (120 Hz), though too fast for the human eye to perceive.

Answer

16.7 ms

If your heart rate is 150 beats per minute during strenuous exercise, what is the time per beat in units of seconds?

Strategy

The heart rate is given as a frequency (150 beats per minute). We need to find the period (time per beat) in seconds. First, convert the frequency to beats per second, then use $T = \frac{1}{f}$ to find the period.

Solution

Convert the heart rate from beats per minute to beats per second:

$$f = 150 \frac{\text{beats}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} = 2.50 \frac{\text{beats}}{\text{s}} = 2.50 \text{ Hz}$$

Now find the period using $T = \frac{1}{f}$:

$$T = \frac{1}{f} = \frac{1}{2.50 \text{ Hz}} = 0.400 \text{ s/beat}$$

Discussion

A period of 0.400 s per beat is reasonable for strenuous exercise. At rest, a typical heart rate might be 60-70 beats per minute (about 1 beat per second), so the higher rate during exercise corresponds to a shorter period between beats. This faster rate allows the heart to pump more blood to supply oxygen to working muscles.

Answer

0.400 s/beat

Find the frequency of a tuning fork that takes $2.50 \times 10^{-3} \text{s}$ to complete one oscillation.

Strategy

The period $T$ is given as the time to complete one oscillation. Use the relationship $f = \frac{1}{T}$ to find the frequency.

Solution

Substitute the given period into $f = \frac{1}{T}$:

$$f = \frac{1}{T} = \frac{1}{2.50 \times 10^{-3} \text{ s}} = \frac{1}{0.00250 \text{ s}} = 400 \text{ Hz}$$

Discussion

A frequency of 400 Hz falls within the range of human hearing (roughly 20 Hz to 20,000 Hz) and corresponds to a musical note near G4 or A4, depending on the tuning standard. Tuning forks are designed to produce pure tones at specific frequencies and are commonly used to tune musical instruments. The period of 2.50 ms is quite short, meaning the fork completes 400 complete vibrations every second.

Answer

400 Hz

A stroboscope is set to flash every $8.00 \times 10^{-5} \text{s}$ . What is the frequency of the flashes?

Strategy

The time between flashes is the period $T$. Use $f = \frac{1}{T}$ to calculate the frequency of the flashes.

Solution

Substitute the given period into $f = \frac{1}{T}$:

$$f = \frac{1}{T} = \frac{1}{8.00 \times 10^{-5} \text{ s}} = \frac{1}{0.0000800 \text{ s}} = 12\,500 \text{ Hz} = 12.5 \text{ kHz}$$

Discussion

The frequency of 12,500 Hz (12.5 kHz) is quite high and actually exceeds the upper limit of human hearing for most people. Stroboscopes are used to make rapidly moving or rotating objects appear stationary by synchronizing the flash rate with the object’s motion. This high frequency allows the stroboscope to “freeze” very fast-moving objects, making 12,500 flashes per second. The period of 80 microseconds is extremely short, demonstrating the rapid nature of the flashing.

Answer

12,500 Hz or 12.5 kHz

A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s?

Strategy

The frequency is the number of vibrations per second. Since each crevice causes one vibration, we need to find how many crevices pass a given point per second. This equals the velocity divided by the spacing between crevices: $f = \frac{v}{d}$.

Solution

Convert the crevice spacing to meters:

$$d = 2.00 \text{ cm} = 0.0200 \text{ m}$$

Calculate the frequency using $f = \frac{v}{d}$:

$$f = \frac{v}{d} = \frac{30.0 \text{ m/s}}{0.0200 \text{ m}} = 1500 \text{ Hz} = 1.50 \text{ kHz}$$

Discussion

A frequency of 1,500 Hz is well within the range of human hearing and would produce an audible tone. This is why tires can make a humming or whining sound at high speeds—the regular pattern of the tread creates vibrations at specific frequencies. The speed of 30.0 m/s is about 108 km/h (67 mph), a typical highway speed. At slower speeds, the frequency would be lower and the pitch of the sound would be deeper. At faster speeds, the pitch would be higher.

Answer

1.50 kHz

Engineering Application

Each piston of an engine makes a sharp sound every other revolution of the engine. (a) How fast is a race car going if its eight-cylinder engine emits a sound of frequency 750 Hz, given that the engine makes 2000 revolutions per kilometer? (b) At how many revolutions per minute is the engine rotating?

Strategy

(a) The sound frequency is determined by how many piston firings occur per second. Since there are 8 cylinders and each fires once every 2 revolutions, we can find the engine’s revolution rate from the sound frequency. Then, using the given relationship between revolutions and distance (2000 rev/km), we can find the car’s speed.

(b) Once we know the engine’s revolution rate in rev/s from part (a), we can convert this to rev/min.

Solution for (a)

First, find how many piston firings occur per engine revolution. With 8 cylinders and each firing once every 2 revolutions:

$$\text{Firings per revolution} = \frac{8 \text{ cylinders}}{2 \text{ rev}} = 4 \text{ firings/rev}$$

The sound frequency equals the number of firings per second, so we can find the engine’s revolution rate:

$$f = 750 \text{ Hz} = 750 \text{ firings/s}$$
$$\text{Rev rate} = \frac{750 \text{ firings/s}}{4 \text{ firings/rev}} = 187.5 \text{ rev/s}$$

Now use the given relationship that the engine makes 2000 revolutions per kilometer to find the speed. Convert km to m:

$$2000 \text{ rev/km} = \frac{2000 \text{ rev}}{1000 \text{ m}} = 2.00 \text{ rev/m}$$

The car’s speed is:

$$v = \frac{187.5 \text{ rev/s}}{2.00 \text{ rev/m}} = 93.8 \text{ m/s}$$

Solution for (b)

Convert the engine’s revolution rate from rev/s to rev/min:

$$\text{Rev rate} = 187.5 \text{ rev/s} \times \frac{60 \text{ s}}{1 \text{ min}} = 11\,250 \text{ rev/min} = 11.3 \times 10^{3} \text{ rev/min}$$

Discussion

The speed of 93.8 m/s is equivalent to about 338 km/h or 210 mph, which is reasonable for a race car at high speed. The engine speed of 11,250 rpm is also typical for a high-performance racing engine under heavy load. The high-pitched 750 Hz sound (close to the musical note G5) is characteristic of racing engines. Note that the relationship between engine revolutions and distance traveled depends on the gear ratio and wheel size—the given value of 2000 rev/km corresponds to a specific gear selection.

Answer

(a) 93.8 m/s

(b) $11.3 \times 10^{3} \text{ rev/min}$

Glossary

period
time it takes to complete one oscillation
periodic motion
motion that repeats itself at regular time intervals
frequency
number of events per unit of time