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Physics: Waves, Sound, and Light - Sound Waves, Hearing, Resonance, and Natural Frequency

Grade 8IB

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

🔑Concepts

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Sound is a longitudinal wave produced by vibrations that requires a medium (solid, liquid, or gas) to travel through. It cannot travel through a vacuum.

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Sound waves consist of Compressions (regions of high pressure where particles are close together) and Rarefactions (regions of low pressure where particles are spread out).

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The speed of sound depends on the medium: it is fastest in solids and slowest in gases because particles in solids are more tightly packed, allowing vibrations to transfer more quickly (vsolid>vliquid>vgasv_{solid} > v_{liquid} > v_{gas}).

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Pitch is directly proportional to the frequency (ff) of the wave. A higher frequency results in a higher pitch.

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Loudness is determined by the amplitude (AA) of the wave. A larger amplitude means more energy and a louder sound, typically measured in decibels (dBdB).

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The human hearing range is typically between 20 Hz20\text{ Hz} and 20,000 Hz20,000\text{ Hz}. Frequencies above this range are called ultrasound, and below are called infrasound.

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Natural Frequency is the frequency at which an object tends to vibrate when disturbed. Resonance occurs when the frequency of an external periodic force matches the natural frequency of an object, leading to a dramatic increase in amplitude.

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The Ear: The outer ear (pinna) collects sound, the middle ear (ossicles) amplifies it, and the inner ear (cochlea) converts vibrations into electrical signals for the brain.

📐Formulae

v=fλv = f \lambda

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

v=dtv = \frac{d}{t}

d=v×t2 (for echo distance to an object)d = \frac{v \times t}{2} \text{ (for echo distance to an object)}

💡Examples

Problem 1:

A student stands 170 m170\text{ m} away from a tall wall and claps their hands. If the speed of sound in air is 340 m/s340\text{ m/s}, calculate the time taken for the student to hear the echo.

Solution:

Using the formula for speed: v=2dtv = \frac{2d}{t}. Rearranging for time: t=2dvt = \frac{2d}{v}. Substituting the values: t=2×170340=340340=1 st = \frac{2 \times 170}{340} = \frac{340}{340} = 1\text{ s}.

Explanation:

In an echo problem, the sound must travel to the wall and back, so the total distance is 2d2d. The time taken is 11 second.

Problem 2:

A tuning fork produces a sound wave with a frequency of 512 Hz512\text{ Hz}. If the wavelength of the sound is 0.65 m0.65\text{ m}, calculate the speed of sound.

Solution:

Using the wave equation: v=fλv = f \lambda. Given f=512 Hzf = 512\text{ Hz} and λ=0.65 m\lambda = 0.65\text{ m}. v=512×0.65=332.8 m/sv = 512 \times 0.65 = 332.8\text{ m/s}.

Explanation:

The speed of a wave is the product of its frequency and its wavelength.

Problem 3:

A string vibrates with a period (TT) of 0.002 s0.002\text{ s}. Determine its frequency and state if it is audible to humans.

Solution:

Using the formula f=1Tf = \frac{1}{T}: f=10.002=500 Hzf = \frac{1}{0.002} = 500\text{ Hz}.

Explanation:

Since 500 Hz500\text{ Hz} falls within the range of 20 Hz20\text{ Hz} to 20,000 Hz20,000\text{ Hz}, the sound is audible to humans.