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Physics - Properties of waves (Light and Sound)

Grade 9IGCSE

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

🔑Concepts

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Waves are oscillations that transfer energy and information without the transfer of matter. They can be categorized as transverse or longitudinal.

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Transverse waves: The oscillations are perpendicular to the direction of energy transfer (e.g., light waves, water waves).

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Longitudinal waves: The oscillations are parallel to the direction of energy transfer, consisting of compressions and rarefactions (e.g., sound waves).

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Wave properties: Amplitude (AA) is the maximum displacement from the equilibrium; Wavelength (λ\lambda) is the distance between two consecutive peaks; Frequency (ff) is the number of waves passing a point per second measured in Hertz (Hz\text{Hz}).

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Reflection: The law of reflection states that the angle of incidence (ii) is equal to the angle of reflection (rr), measured from the normal.

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Refraction: When a wave enters a medium of different optical density, its speed changes, causing it to bend. Light bends towards the normal when entering a denser medium (n1<n2n_1 < n_2).

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Total Internal Reflection (TIR): Occurs when light travels from a more dense medium to a less dense medium and the angle of incidence is greater than the critical angle (cc).

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Sound properties: Sound requires a medium to travel (cannot travel in a vacuum). The pitch is determined by frequency (ff), and loudness is determined by amplitude (AA).

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The speed of light in a vacuum is approximately c=3.0×108 m/sc = 3.0 \times 10^8 \text{ m/s}, while the speed of sound in air is approximately 330–340 m/s330 \text{--} 340 \text{ m/s}.

📐Formulae

v=fλv = f \lambda

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

n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}

n=cvn = \frac{c}{v}

sin⁡c=1n\sin c = \frac{1}{n}

💡Examples

Problem 1:

A water wave has a wavelength of 0.05 m0.05\text{ m} and a frequency of 10 Hz10\text{ Hz}. Calculate the speed of the wave.

Solution:

v=fλ=10×0.05=0.5 m/sv = f \lambda = 10 \times 0.05 = 0.5\text{ m/s}

Explanation:

The wave speed is calculated by multiplying the frequency by the wavelength using the standard wave equation.

Problem 2:

Calculate the critical angle cc for a glass block with a refractive index of n=1.52n = 1.52.

Solution:

sin⁡c=1n=11.52≈0.6579\sin c = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579 c=sin⁡−1(0.6579)≈41.1∘c = \sin^{-1}(0.6579) \approx 41.1^\circ

Explanation:

To find the critical angle, we use the relationship between the refractive index and the sine of the critical angle, then take the inverse sine.

Problem 3:

An echo is heard 2.0 s2.0\text{ s} after a sound is made near a large wall. If the speed of sound is 340 m/s340\text{ m/s}, how far away is the wall?

Solution:

v=2dt  ⟹  d=v×t2v = \frac{2d}{t} \implies d = \frac{v \times t}{2} d=340×2.02=340 md = \frac{340 \times 2.0}{2} = 340\text{ m}

Explanation:

For echo problems, the sound travels to the wall and back, covering a total distance of 2d2d. We divide by 22 to find the distance to the wall.