Wave Speed Calculator

Solve v = fλ for wave speed, and find the speed of a wave on a string, step by step
A wave has a frequency of 250 Hz and a wavelength of 1.4 m. Find its speed.
Find the wavelength of a 512 Hz tone in air at 343 m/s
A string under 80 N tension has linear density 0.0050 kg/m. Find the wave speed.
A wave of period 0.02 s and wavelength 3 m — what is its speed?

The Wave Speed Formula

A wave advances one wavelength in each period, so its speed is wavelength times frequency:

v=fλv = f\lambda

Symbols and SI units:

  • vv — wave speed, metres per second (m/s)
  • ff — frequency, hertz (Hz), i.e. cycles per second
  • λ\lambda — wavelength, metres (m)

Because the period is T=1/fT = 1/f, the same relation can be written v=λ/Tv = \lambda/T — useful when a problem quotes a period in seconds instead of a frequency.

When it applies: to any periodic travelling wave — sound, light, water ripples, waves on a rope.

The key idea that trips people up: the speed is set by the medium, not by the source. Turning a tuning fork up in frequency does not speed the sound up; the wavelength shortens instead, keeping fλf\lambda constant. The frequency is what the source fixes.

The assumption people forget: v=fλv = f\lambda describes a wave in a single uniform medium. Cross into a new medium and vv and λ\lambda both change while ff stays put.

Where the Speed Comes From

If the wave speed is not given, it comes from a property of the medium.

Transverse wave on a string:

v=Tμv = \sqrt{\frac{T}{\mu}}

with TT the tension in newtons (N) and μ\mu the linear mass density in kilograms per metre (kg/m). Tightening a guitar string raises vv, and with λ\lambda fixed by the string length, the pitch rises.

Sound in air: about 343343 m/s at 2020 °C, rising roughly 0.60.6 m/s per °C.

Light in vacuum: c=2.998×108c = 2.998 \times 10^8 m/s; in a medium of refractive index nn, v=c/nv = c/n.

Rearranged, the same formula gives the other two unknowns:

f=vλ,λ=vff = \frac{v}{\lambda}, \qquad \lambda = \frac{v}{f}

The assumption people forget: μ\mu is mass per unit length, not a density in kg/m³. Divide a string's total mass by its total length.

Common Mistakes to Avoid

  • Using 3×1083 \times 10^8 m/s for sound — sound in air is about 343343 m/s, six orders of magnitude slower.
  • Believing a louder or higher-pitched wave travels faster — amplitude and frequency do not change the speed; only the medium does.
  • Leaving prefixes unconverted2.42.4 GHz is 2.4×1092.4 \times 10^9 Hz and 600600 nm is 6.0×1076.0 \times 10^{-7} m. Convert before multiplying.
  • Confusing frequency with period — they are reciprocals, so a 0.020.02 s period is 5050 Hz, not 0.020.02 Hz.
  • Using kg/m³ in v=T/μv = \sqrt{T/\mu} — that formula needs linear density in kg/m.
  • Assuming the wavelength is fixed when a wave enters a new medium — the frequency is what carries over unchanged.

示例题目

Step 1: v=fλv = f\lambda
Step 2: v=(250 Hz)(1.4 m)=(250 s1)(1.4 m)v = (250\ \text{Hz})(1.4\ \text{m}) = (250\ \text{s}^{-1})(1.4\ \text{m})
Step 3: v=350 m/sv = 350\ \text{m/s}
Answer: v=350v = 350 m/s

Step 1: Rearrange v=fλv = f\lambda to λ=v/f\lambda = v/f
Step 2: λ=(343 m/s)÷(512 s1)\lambda = (343\ \text{m/s}) \div (512\ \text{s}^{-1})
Step 3: λ=0.670 m\lambda = 0.670\ \text{m}
Answer: λ0.670\lambda \approx 0.670 m (about 6767 cm)

Step 1: v=T/μ=(80 N)÷(0.0050 kg/m)v = \sqrt{T/\mu} = \sqrt{(80\ \text{N}) \div (0.0050\ \text{kg/m})}
Step 2: v=16000 m2/s2=126.5 m/sv = \sqrt{16\,000\ \text{m}^2/\text{s}^2} = 126.5\ \text{m/s}
Step 3: Then f=v/λ=(126.5 m/s)÷(1.2 m)f = v/\lambda = (126.5\ \text{m/s}) \div (1.2\ \text{m})
Step 4: f=105 Hzf = 105\ \text{Hz}
Answer: v126.5v \approx 126.5 m/s and f105f \approx 105 Hz

常见问题

v = fλ — the speed in metres per second equals the frequency in hertz times the wavelength in metres. Equivalently v = λ/T, since the period T is the reciprocal of the frequency.

Read the wavelength off a displacement-versus-position graph and the period off a displacement-versus-time graph, then compute v = λ/T. Using one graph alone is not enough, because each shows only half the information.

In a non-dispersive medium, no. The speed is set by the medium, so raising the frequency shortens the wavelength and leaves fλ unchanged. Dispersive media such as glass are the exception, which is why a prism splits white light.

Use v = √(T/μ), where T is the tension in newtons and μ is the mass per unit length in kg/m. An 80 N tension on a 0.0050 kg/m string gives √16,000 = 126.5 m/s.

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