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Interactive physics simulator

Pitch and Frequency

Investigate how sound pitch corresponds to frequency. Slide the bridge of a real-world Monochord string to alter its vibrating length, generate acoustic waves to inspect period cycles on a virtual oscilloscope, and test the biological hearing ranges of humans and animals.

Acoustic Pitch Laboratory ⓘ

Interact with vibrating strings, particle wave generators, and audibility limiters. Enable the Audio Synth checkbox to hear the pitch.

Monochord Lab

Live Acoustics Telemetry

Source Vibration
Monochord String
Fundamental Frequency
187.5 Hz
Wave Period (T)
5.33 ms
Vibrating Length (L)
0.80 m
String Tension (T)
360 N
Acoustic Regime
Audible Sound

The Physics of Pitch

Pitch is the subjective sensory perception of the frequency of a sound wave. When an object vibrates, it creates pressure oscillations in the surrounding air. The rate at which these oscillations repeat determines the wave's frequency. The human brain translates higher frequencies into high-pitched sounds, and lower frequencies into low-pitched sounds.

While pitch is closely correlated with frequency, they are not identical: frequency is an objective physical quantity measured in Hertz (Hz, cycles per second), whereas pitch is a physiological and psychological response of the auditory system.

The Monochord and laws of Vibrating Strings

A monochord is an ancient laboratory instrument consisting of a single string tensioned over a sound box with a movable bridge. The bridge divides the string into two vibrating segments. Clamping the string at both ends forces these boundaries to remain stationary, forming displacement nodes, while the center of the string reaches maximum displacement, forming an antinode.

The fundamental wavelength of the standing wave formed on a string of length L is:

λ = 2L

The speed of a transverse wave traveling on a string depends on the tension (T) and the linear mass density (μ):

v = √(T / μ)

Combining these equations yields the fundamental frequency (Mersenne's Law) of a vibrating string:

f1 = v / λ = (1 / 2L) · √(T / μ)

This formula leads to the three fundamental laws of vibrating strings:

  • Law of Length: Frequency is inversely proportional to string length (f ∝ 1/L). Halving the length doubles the frequency (up one octave).
  • Law of Tension: Frequency is directly proportional to the square root of tension (f ∝ √T). Quadrupling the tension doubles the frequency.
  • Law of Mass: Frequency is inversely proportional to the square root of the linear density (f ∝ 1/√μ). A thicker, heavier string vibrates slower, producing a deeper pitch.

Infrasound, Audible Range, and Ultrasound

The acoustic spectrum is categorized based on human audibility:

Regime Frequency Range Characteristics & Physics Examples
Infrasound f < 20 Hz Low-frequency mechanical vibrations. Travels extremely long distances with minimal absorption through ground and air. Earthquake seismic waves, volcanic eruptions, elephant communications.
Audible Sound 20 Hz ≤ f ≤ 20,000 Hz Frequencies detectable by the stereocilia inside the human cochlea. Lower limit is 20 Hz, upper limit decreases with age. Human speech, musical instruments, birdsong.
Ultrasound f > 20,000 Hz High-frequency mechanical oscillations. Due to short wavelengths, they scatter less and can be focused into tight beams. Bat sonar navigation, medical sonography, dog whistles, marine sonar.

📝 Solved Examples

Example 1

A steel monochord wire of length 80.0 cm has a linear mass density of 4.00 g/m. If the wire is tightened to a tension of 360 N, calculate: (a) the velocity of the transverse wave on the string, and (b) the fundamental frequency (pitch) produced when the string is plucked.

Step-by-step Solution

Step 1: Identify given parameters in SI units:
Length (L) = 80.0 cm = 0.80 m
Linear mass density (μ) = 4.00 g/m = 0.0040 kg/m
Tension (T) = 360 N

Step 2: Calculate the wave velocity (v) using the formula v = √(T / μ):
v = √(360 N / 0.0040 kg/m) = √(90,000) = 300 m/s

Step 3: Calculate the fundamental frequency (f₁) using f₁ = v / 2L:
f₁ = 300 m/s / (2 · 0.80 m) = 300 / 1.6 = 187.5 Hz.
(This is the perceived pitch, lying close to the F₃ musical note).

Answer: (a) Wave speed = 300 m/s, (b) Fundamental Pitch = 187.5 Hz

Example 2

An observer captures a pure acoustic tone on an oscilloscope. The time division dial is set to 0.50 milliseconds per grid square. The displayed sinusoidal wave cycle spans exactly 4.0 horizontal grid squares from one crest to the next. Calculate the period of the sound wave and its frequency, and state whether it is audible to a normal human ear.

Step-by-step Solution

Step 1: Calculate the period (T) of the wave by multiplying the horizontal divisions by the time division setting:
Period (T) = 4.0 divisions · 0.50 ms/division = 2.0 ms = 0.0020 seconds.

Step 2: Calculate the frequency (f) using the reciprocal of the period (f = 1 / T):
f = 1 / 0.0020 s = 500 Hz.

Step 3: Compare the frequency with the human audibility range (20 Hz to 20,000 Hz). Since 500 Hz falls within this range, it is audible as a mid-pitched tone.

Answer: Period = 2.0 ms, Frequency = 500 Hz (Audible)

Example 3

Explain how the pitch of a vibrating guitar string changes if the player: (a) slides their finger down the fretboard to halve the vibrating string length, and (b) adjusts the tuning peg to double the tension of the string.

Step-by-step Solution

Step 1: Express the fundamental frequency equation: f₁ = (1 / 2L) · √(T / μ).

Step 2: Analyze length change (a). Since frequency is inversely proportional to length (f ∝ 1/L), halving the length (L' = L / 2) doubles the frequency: f' = 2 · f₁. The pitch rises by exactly one octave.

Step 3: Analyze tension change (b). Since frequency is directly proportional to the square root of tension (f ∝ √T), doubling the tension (T' = 2T) increases the frequency by a factor of √2 ≈ 1.414: f' = 1.414 · f₁. The pitch rises by about 6 semitones.

Answer: (a) Frequency doubles (up one octave), (b) Frequency increases by √2 (~41.4% increase)

💡 Concept Check & Practice

Q1. What is the physical difference between the loudness and the pitch of a sound wave?

Loudness is a subjective sensation determined by the amplitude (or intensity) of the sound wave, which is related to the energy carried by the wave. A larger amplitude represents a louder sound. Pitch is a subjective sensation determined by the frequency (or period) of the wave. A higher frequency represents a higher-pitched sound. Adjusting the volume of a speaker changes its amplitude and loudness, but does not alter its frequency or pitch.

Q2. How does a monochord bridge slide affect the wavelength and frequency of the standing wave produced?

A monochord string clamped at both ends produces standing waves where the ends are displacement nodes. The fundamental wavelength is given by λ = 2L, where L is the vibrating length between the fixed end and the bridge. When the bridge is slid toward the fixed end (shortening L), the wavelength is shortened. Because the wave speed (v) on the string remains constant (since tension and density are unchanged), the frequency must increase (f = v / λ = v / 2L), resulting in a higher perceived pitch.

Q3. Why does the sound of a whistle have a high pitch while a thunderclap has a low pitch, in terms of molecular vibration?

The sound of a whistle is caused by air blowing across a small cavity, forcing air molecules to compress and expand at a rapid rate, resulting in thousands of vibrations per second (high frequency). This rapid oscillation is perceived as a high pitch. A thunderclap is produced by the rapid expansion of air heated by a lightning bolt, creating a massive displacement of air that vibrates slowly over a larger volume, resulting in a low frequency (often below 100 Hz). This slow oscillation is perceived as a low-pitched rumble.

Q4. Define infrasound and ultrasound, and describe one biological or technological application for each.

Infrasound refers to sound waves with frequencies below 20 Hz, which is below the human limit of hearing. Elephants use infrasound for long-distance communication through ground vibrations. Ultrasound refers to sound waves with frequencies above 20,000 Hz (20 kHz), which is above the human limit. Bats use ultrasound (echolocation) to navigate and hunt in the dark by emitting high-frequency squeaks and measuring the echo delays.

Frequently Asked Questions

What is pitch in sound?
Pitch is how high or low a sound seems to the human ear. In physics, pitch mainly depends on the frequency of the sound wave.
What is the formula related to pitch and frequency?
For a wave, the basic relation is v = fλ. On a string fixed at both ends, the fundamental frequency is f1 = v / 2L.
Does higher frequency mean higher pitch?
Yes. A higher frequency means more vibrations per second, so the sound is heard as a higher pitch.
Does amplitude change pitch?
No. Amplitude mainly changes loudness. Pitch depends mainly on frequency, not on how large the wave is.
How does string length affect pitch?
A shorter vibrating string gives a higher frequency and therefore a higher pitch. A longer string gives a lower frequency and a lower pitch.
How does string tension affect pitch?
Increasing tension makes waves travel faster along the string, which increases frequency and raises pitch.
What is infrasound?
Infrasound is sound below 20 Hz. It is usually too low for humans to hear.
What is ultrasound?
Ultrasound is sound above 20,000 Hz. Humans normally cannot hear it, but animals like bats and dolphins can use it.
Why do small instruments often sound higher pitched?
Small instruments often have shorter vibrating strings, smaller air columns, or smaller membranes, so they tend to vibrate faster and produce higher frequencies.
How does the pitch simulator work?
The simulator changes string length, tension, material, or generated frequency and then updates the wave period, frequency, and hearing range values live.