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Physics - Sound: Vibrations and Pitch

Grade 7IGCSE

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

🔑Concepts

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

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.

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

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).

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

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

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.

Sound: Vibrations and Pitch - Revision Notes & Key Formulas | IGCSE Grade 7 Science