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Physics: Waves, Sound, and Light - Wave Equation, Frequency, Wavelength, and Amplitude

Grade 8IB

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

🔑Concepts

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A Wave is a disturbance that transfers energy from one location to another without transferring matter.

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Amplitude (AA) is the maximum displacement of a point on the wave from its undisturbed (rest) position. Larger amplitude corresponds to louder sound or brighter light.

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Wavelength (λ\lambda) is the distance between two consecutive corresponding points on a wave, such as from crest to crest or trough to trough, measured in meters (mm).

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Frequency (ff) is the number of complete waves passing a fixed point per second, measured in Hertz (HzHz). Higher frequency corresponds to a higher pitch in sound.

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Period (TT) is the time taken (in seconds) for one complete wave cycle to pass a point. It is the reciprocal of frequency: T=1fT = \frac{1}{f}.

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Wave Speed (vv) is the speed at which energy is transferred through a medium. For any wave, speed is the product of its frequency and wavelength.

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Transverse Waves (e.g., light) have oscillations perpendicular to the direction of energy transfer, while Longitudinal Waves (e.g., sound) have oscillations parallel to the direction of energy transfer.

📐Formulae

v=fλv = f \lambda

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

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

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

💡Examples

Problem 1:

A tuning fork produces a sound wave with a frequency of 256 Hz256\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=340256≈1.33 m\lambda = \frac{v}{f} = \frac{340}{256} \approx 1.33\text{ m}

Explanation:

Using the wave equation v=fλv = f \lambda, we rearrange it to solve for wavelength: λ=vf\lambda = \frac{v}{f}. Substituting the given values v=340v = 340 and f=256f = 256 gives the result.

Problem 2:

A wave on a string has a period of 0.02 s0.02\text{ s} and a wavelength of 0.5 m0.5\text{ m}. Find the wave speed.

Solution:

f=1T=10.02=50 Hzf = \frac{1}{T} = \frac{1}{0.02} = 50\text{ Hz} v=fλ=50×0.5=25 m/sv = f \lambda = 50 \times 0.5 = 25\text{ m/s}

Explanation:

First, calculate the frequency ff from the period TT using f=1Tf = \frac{1}{T}. Then, use the wave equation v=fλv = f \lambda to find the speed.

Problem 3:

A wave travels 120 meters120\text{ meters} in 3 seconds3\text{ seconds}. If its frequency is 10 Hz10\text{ Hz}, what is its wavelength?

Solution:

v=dt=1203=40 m/sv = \frac{d}{t} = \frac{120}{3} = 40\text{ m/s} λ=vf=4010=4 m\lambda = \frac{v}{f} = \frac{40}{10} = 4\text{ m}

Explanation:

First, calculate the wave speed vv using the distance-time formula. Then, substitute vv and the frequency ff into the rearranged wave equation λ=vf\lambda = \frac{v}{f}.