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Waves - Properties of Sound

Grade 9IB

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

🔑Concepts

Sound is a longitudinal mechanical wave that requires a medium (solid, liquid, or gas) to travel; it cannot travel through a vacuum.

Sound travels through a series of compressions (high pressure) and rarefactions (low pressure) in the medium.

The pitch of a sound is determined by its frequency (ff). A high-frequency sound has a high pitch, while a low-frequency sound has a low pitch.

The loudness of a sound is determined by its amplitude (AA). A larger amplitude means more energy and a louder sound.

The human hearing range is approximately 20 Hz20\text{ Hz} to 20,000 Hz20,000\text{ Hz}. Sounds above 20 kHz20\text{ kHz} are known as ultrasound.

The speed of sound (vv) depends on the medium: it travels fastest in solids, then liquids, and slowest in gases (approx. 340 m/s340\text{ m/s} in air at room temperature).

Echoes are the reflection of sound waves off a surface. For a clear echo, the reflecting surface must be at a sufficient distance to distinguish the original sound from the reflected one.

📐Formulae

v=fλv = f \lambda

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

v=2dt (for echoes/reflection)v = \frac{2d}{t} \text{ (for echoes/reflection)}

Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

💡Examples

Problem 1:

A musical note has a frequency of 440 Hz440\text{ Hz}. If the speed of sound in air is 340 m/s340\text{ m/s}, calculate the wavelength of the sound wave.

Solution:

λ=vf=340 m/s440 Hz0.77 m\lambda = \frac{v}{f} = \frac{340\text{ m/s}}{440\text{ Hz}} \approx 0.77\text{ m}

Explanation:

Using the wave equation v=fλv = f\lambda, we rearrange to solve for wavelength λ\lambda. Dividing the speed of sound by the frequency gives the distance between consecutive compressions.

Problem 2:

A ship sends an ultrasound pulse to the seabed to determine the depth of the ocean. The pulse is reflected back and detected 1.2 seconds1.2\text{ seconds} later. If the speed of sound in seawater is 1500 m/s1500\text{ m/s}, what is the depth of the ocean?

Solution:

d=v×t2=1500 m/s×1.2 s2=900 md = \frac{v \times t}{2} = \frac{1500\text{ m/s} \times 1.2\text{ s}}{2} = 900\text{ m}

Explanation:

Since the sound travels to the seabed and back, the total distance covered is 2d2d. We use the formula 2d=v×t2d = v \times t and divide by 22 to find the one-way distance (depth).

Problem 3:

Calculate the period (TT) of a sound wave that has a frequency of 2 kHz2\text{ kHz}.

Solution:

T=1f=12000 Hz=0.0005 s=0.5 msT = \frac{1}{f} = \frac{1}{2000\text{ Hz}} = 0.0005\text{ s} = 0.5\text{ ms}

Explanation:

The period is the reciprocal of the frequency. First, convert 2 kHz2\text{ kHz} to 2000 Hz2000\text{ Hz}, then apply the formula T=1fT = \frac{1}{f}.