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New YorkPhysicsSyllabus dot point

What quantities describe a wave, and how are wave speed, frequency and wavelength related?

Define amplitude, wavelength, frequency and period, distinguish transverse and longitudinal waves, and apply the wave equation v=fλv = f\lambda and the period-frequency relationship T=1/fT = 1/f.

A Regents Physics answer on wave properties and the wave equation: amplitude, wavelength, frequency and period, transverse versus longitudinal waves, and the Reference-Table equations linking wave speed, frequency and wavelength, with worked examples.

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  1. What this topic is asking
  2. The properties of a wave
  3. Transverse and longitudinal waves
  4. The wave equation
  5. Reference Tables note
  6. Try this

What this topic is asking

This dot point opens the waves strand by defining the quantities that describe any wave and linking them with the wave equation. The Physical Setting/Physics course asks you to define amplitude, wavelength, frequency and period, to distinguish transverse from longitudinal waves, and to apply v=fλv = f\lambda and T=1fT = \dfrac{1}{f}. The Regents tests these as definitions, diagram-reading, and calculations linking speed, frequency and wavelength.

The properties of a wave

These quantities describe any periodic wave. The amplitude relates to the energy the wave carries (a louder sound or brighter light has a larger amplitude), while the frequency relates to pitch for sound or color for light. Frequency and period are reciprocals, so a higher frequency means a shorter period. Reading these off a wave diagram, distinguishing wavelength (a distance) from period (a time), is a common Regents skill.

Transverse and longitudinal waves

The distinction matters because some behaviors (like polarization) occur only for transverse waves. On the Regents, light and other electromagnetic waves are always transverse and can travel through a vacuum, while sound is longitudinal and needs a medium (it cannot travel through a vacuum).

The wave equation

The wave equation ties the three central quantities together: any one can be found from the other two. A key consequence is that in a given medium the wave speed is fixed, so frequency and wavelength are inversely related: a higher-frequency wave has a shorter wavelength. The frequency is set by the source and does not change when a wave passes into a new medium; the speed and wavelength change together to keep v=fλv = f\lambda satisfied (the basis of refraction).

Reference Tables note

The Reference Tables print the wave equation v=fλv = f\lambda and the period-frequency relationship T=1fT = \dfrac{1}{f} in the Waves section. The definitions of amplitude, wavelength, frequency and period, and the transverse-longitudinal distinction, are recall items. The speed of light c=3.00×108c = 3.00 \times 10^8 m/s in the constants list is the wave speed for all electromagnetic waves in a vacuum, used with v=fλv = f\lambda for the electromagnetic spectrum.

Try this

Q1. State the difference between a transverse and a longitudinal wave. [2 points]

  • Cue. Transverse: the medium vibrates perpendicular to the wave direction. Longitudinal: the medium vibrates parallel to the wave direction.

Q2. A wave has a frequency of 8.08.0 Hz and wavelength 0.500.50 m. Calculate its speed. [2 points]

  • Cue. v=fλ=(8.0)(0.50)=4.0v = f\lambda = (8.0)(0.50) = 4.0 m/s.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). A wave has a frequency of 5.05.0 Hz and a wavelength of 0.400.40 m. Calculate the speed of the wave. Show the equation, substitution and answer.
Show worked answer →

A 2-point constructed-response calculation using the Reference-Table wave equation v=fλv = f\lambda.

Equation: v=fλv = f\lambda.
Substitution: v=(5.0)(0.40)v = (5.0)(0.40).
Answer: v=2.0v = 2.0 m/s.

Markers reward the equation from the tables, correct substitution with units, and the speed in m/s. A common error is dividing rather than multiplying the frequency and wavelength.

Regents (style)2 marksPart B-2 (constructed response). A sound wave travels at 340340 m/s and has a frequency of 170170 Hz. Calculate its wavelength. Show the equation, substitution and answer.
Show worked answer →

A 2-point constructed-response calculation rearranging the wave equation.

Equation: v=fλv = f\lambda, rearranged to λ=vf\lambda = \dfrac{v}{f}.
Substitution: λ=340170\lambda = \dfrac{340}{170}.
Answer: λ=2.0\lambda = 2.0 m.

Markers reward the rearranged equation, correct substitution with units, and the wavelength in meters. The wave equation links all three quantities, so any one can be found from the other two.

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