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Physics - Sound: Vibrations and Pitch

Grade 7Cambridge (IGCSE)

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Sound is a form of energy produced by vibrating objects. These vibrations travel through a medium (solid, liquid, or gas) as longitudinal waves.

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The pitch of a sound refers to how 'high' or 'low' it sounds to the ear. It is directly determined by the frequency of the vibration.

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Frequency (ff) is the number of complete vibrations or oscillations per second. It is measured in Hertz (HzHz).

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A high-frequency vibration produces a high-pitched sound (e.g., a whistle), while a low-frequency vibration produces a low-pitched sound (e.g., a bass drum).

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The time period (TT) is the time taken for one complete vibration. It is the inverse of frequency: T=1fT = \frac{1}{f}.

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Amplitude is the maximum displacement of a particle from its rest position. While frequency determines pitch, amplitude determines the loudness of the sound.

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The human hearing range is typically between 20 Hz20\ Hz and 20,000 Hz20,000\ Hz (or 20 kHz20\ kHz). Sounds with frequencies above this range are called ultrasound.

📐Formulae

f=Number of vibrationsTotal time (s)f = \frac{\text{Number of vibrations}}{\text{Total time (s)}}

f=1Tf = \frac{1}{T}

T=1fT = \frac{1}{f}

Unit: 1 Hz=1 vibration per second\text{Unit: } 1\ Hz = 1\ \text{vibration per second}

💡Examples

Problem 1:

A tuning fork vibrates 1,2001,200 times in 33 seconds. Calculate its frequency and state whether its pitch is higher or lower than a tuning fork vibrating at 500 Hz500\ Hz.

Solution:

f=12003=400 Hzf = \frac{1200}{3} = 400\ Hz

Explanation:

The frequency is 400 Hz400\ Hz. Since 400 Hz<500 Hz400\ Hz < 500\ Hz, this tuning fork has a lower pitch.

Problem 2:

An oscilloscope shows that a sound wave has a time period (TT) of 0.005 s0.005\ s. What is the frequency of this sound?

Solution:

f=10.005=200 Hzf = \frac{1}{0.005} = 200\ Hz

Explanation:

By using the reciprocal relationship between period and frequency, we divide 11 by the time in seconds to find the frequency in Hertz.

Problem 3:

A bee beats its wings with a frequency of 250 Hz250\ Hz. How many times do its wings beat in 1010 seconds?

Solution:

Vibrations=f×time=250 Hz×10 s=2,500\text{Vibrations} = f \times \text{time} = 250\ Hz \times 10\ s = 2,500

Explanation:

Frequency tells us the vibrations per second. Multiplying frequency by the total time gives the total number of vibrations.