Pitch depends on the frequency of a sound wave, and loudness depends on its amplitude. They are separate properties: changing one does not have to change the other.
This skill appears in electromagnetic spectrum and sound when a question shows two sound traces on a screen and asks which is higher or louder, or asks for a frequency from a trace.
How do I read a trace?
A microphone turns sound into a voltage, and an oscilloscope displays it against time. The horizontal axis is time, and the vertical axis shows displacement, or voltage.
Two features matter. The height of a peak above the middle line is the amplitude. The time for one complete wave, from one peak to the next, is the period T, and frequency = 1 ÷ T.
A narrower wave on the screen means a shorter period and a higher frequency. A taller wave means a larger amplitude and a louder sound.
Worked example
Two traces, A and B, share one screen and one time scale. (Invented example data.) The scale is 1.0 ms per division and the screen is 10 divisions wide.
Trace A shows 2 complete waves and its peaks are 3.0 divisions high. Trace B shows 4 complete waves and its peaks are 1.5 divisions high. Compare the pitch and loudness.
Step 1, time across the screen: 10 × 1.0 ms = 10 ms.
Step 2, period of A: 10 ms ÷ 2 = 5.0 ms = 0.0050 s. Frequency = 1 ÷ 0.0050 = 200 Hz.
Step 3, period of B: 10 ms ÷ 4 = 2.5 ms = 0.0025 s. Frequency = 1 ÷ 0.0025 = 400 Hz.
Step 4, compare: B has double the frequency, so it has the higher pitch. A has double the amplitude (3.0 against 1.5 divisions), so A is the louder sound.
Step 5, check: 200 × 0.0050 = 1 and 400 × 0.0025 = 1, consistent with f × T = 1.
The mistake to watch for
A common slip is to treat loud and high as the same thing. A student looks at the tall, slow wave in trace A and writes “A has the higher pitch because it is bigger”.
Mistaken answer: A has a higher pitch because the waves are taller.
Height shows amplitude, which relates to loudness, not pitch.
The correction is to ask two separate questions: “how many waves fit across the screen?” for pitch, and “how tall are the waves?” for loudness. Write the two answers on different lines.
Check yourself
1. One complete wave on a trace takes 0.0020 s. Find the frequency.
Show answer
f = 1 ÷ T = 1 ÷ 0.0020 = 500 Hz.
2. A speaker’s volume is turned up but the note stays the same. What changes on the trace?
Show answer
The amplitude increases, so the waves become taller. The frequency and pitch stay the same, so the number of waves across the screen does not change.
3. A bat emits sound at 50 kHz. Can a person hear it? Find the period of the wave.
Show answer
50 kHz = 50 000 Hz, which is above the approximate upper limit of human hearing of 20 000 Hz, so a person cannot hear it. T = 1 ÷ 50 000 = 0.000020 s = 2.0 × 10⁻⁵ s (20 microseconds).
Where this leads next
Now test all five lessons together in the electromagnetic spectrum and sound practice set. To revise the equation behind frequency and period, see wave behaviour.
Some students can calculate frequency yet still mix up which feature means which. A teacher can ask you to sketch both traces from a description, which is the style of online one-to-one Physics tuition.