krit.club logo

Waves - Transverse and Longitudinal Waves

Grade 9IB

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

🔑Concepts

•

A wave is a disturbance that transfers energy from one point to another without the permanent transfer of matter.

•

In a Transverse Wave, the particles of the medium oscillate perpendicular (90∘90^\circ) to the direction of energy transfer. Examples include light waves, radio waves, and water waves.

•

Transverse waves are characterized by peaks called crests and dips called troughs.

•

In a Longitudinal Wave, the particles of the medium oscillate parallel to the direction of energy transfer. The primary example is a sound wave.

•

Longitudinal waves consist of regions of high pressure called compressions and regions of low pressure called rarefactions.

•

Amplitude (AA) is the maximum displacement of a particle from its equilibrium position, measured in meters (mm).

•

Wavelength (λ\lambda) is the distance between two consecutive identical points on a wave (e.g., crest to crest or compression to compression).

•

Frequency (ff) is the number of complete wave cycles passing a point per second, measured in Hertz (HzHz).

•

Period (TT) is the time taken for one complete wave cycle to pass a point, measured in seconds (ss).

📐Formulae

v=fλv = f \lambda

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

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

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

💡Examples

Problem 1:

A sound wave traveling through air has a frequency of 440Hz440 Hz. If the speed of sound is 340m/s340 m/s, calculate the wavelength (λ\lambda) of the sound wave.

Solution:

Given: f=440Hzf = 440 Hz, v=340m/sv = 340 m/s. Using v=fλv = f \lambda, we rearrange to find λ=vf\lambda = \frac{v}{f}. λ=340440≈0.77m\lambda = \frac{340}{440} \approx 0.77 m

Explanation:

To find the wavelength, the wave speed is divided by the frequency. The resulting unit is meters (mm).

Problem 2:

A buoy in the ocean oscillates up and down 1212 times in 6060 seconds as waves pass by. Calculate the period (TT) and the frequency (ff) of the waves.

Solution:

f=cyclestime=1260=0.2Hzf = \frac{\text{cycles}}{\text{time}} = \frac{12}{60} = 0.2 Hz T=1f=10.2=5sT = \frac{1}{f} = \frac{1}{0.2} = 5 s

Explanation:

Frequency is the number of oscillations per unit time. The period is the reciprocal of the frequency, representing the time for one single oscillation.

Problem 3:

Calculate the speed of a transverse wave on a string if the distance between a crest and the adjacent trough is 0.5m0.5 m and the frequency is 10Hz10 Hz.

Solution:

Distance from crest to trough is 12λ\frac{1}{2}\lambda. Therefore, λ=2×0.5m=1.0m\lambda = 2 \times 0.5 m = 1.0 m. Using v=fλv = f \lambda: v=10Hz×1.0m=10m/sv = 10 Hz \times 1.0 m = 10 m/s

Explanation:

The wavelength is the distance of a full cycle (crest to crest). Since crest to trough is half a cycle, we double it before calculating speed.