A Clamped String's Wavelength

solved TravellingWave

Q. The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2π/3 x) $k = 120$ deg/m cos (120 πt) where x and y are in m and t in s. Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength of each wave, in metres?

Outputs

wavelength

λ

How it's solved

Solved with TravellingWave

Given

$k = 120$ deg/m from “sin (2π/3 x)”

Find $λ$ .

Working

$k = 120$ deg/m $= 2.0943951$ m⁻¹
wave number and wavelength(wave) $k = \frac{2 \cdot π}{λ}$
$2.0943951 = \frac{2 \cdot 3.14159265}{λ}$
$λ = 3.00 m$
$λ \approx 3.0$ mto 2 significant figures

Conclusion

The two running waves share the standing wave's wave number, and a full turn of phase along the string is one wavelength.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_clamped_string_s_wavelength do4  kind :numerical5  describes "The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin (2π/3 x) " \6            "cos (120 πt) where x and y are in m and t in s. Interpret the wave as a superposition of two waves " \7            "travelling in opposite directions. %{wavelength}, in metres?"8  asks wavelength: "What is the wavelength of each wave"9end

solution.rb

1require_relative "../questions/a_clamped_string_s_wavelength"2require_relative "../science/waves/travelling_wave"34solve :a_clamped_string_s_wavelength, using: TravellingWave do5  objects :wave67  deduce_from "sin (2π/3 x)", wave.wave_number => "120 deg/m"89  wavelength = wave.wavelength10  answer wavelength:, in: "m", figures: 21112  conclusion "The two running waves share the standing wave's wave number, and a full turn of phase along the " \13             "string is one wavelength."14end

playground.rb

1require_relative "../solutions/a_clamped_string_s_wavelength"23play_with :a_clamped_string_s_wavelength do4  title "A Clamped String's Wavelength"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	TravellingWave
amplitude	$a$	m	—	TravellingWave
wave number	$k$	m⁻¹	—	TravellingWave
angular frequency	$ω$	rad/s	—	TravellingWave
wavelength	$λ$	m	0 … ∞	TravellingWave
period	$T$	s	—	TravellingWave
frequency	$ν$	Hz	0 … ∞	TravellingWave
wave speed	$v$	m/s	—	TravellingWave
separation	$Δ x$	m	—	TravellingWave
phase difference	$Δ φ$	—	—	TravellingWave
wave number of ground	$k_{g}$	m⁻¹	—
wavelength of ground	$λ_{g}$	m	0 … ∞
wave number of wave	$k$	m⁻¹	—
wavelength of wave	$λ$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	—
period of ground	$T_{g}$	s	—
angular frequency of wave	$ω$	rad/s	—
period of wave	$T$	s	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
frequency of wave	$ν$	Hz	0 … ∞
wave speed of ground	$v_{g}$	m/s	—
wave speed of wave	$v$	m/s	—
phase difference of ground	$Δ φ_{g}$	—	—
separation of ground	$Δ x_{g}$	m	—
phase difference of wave	$Δ φ$	—	—
separation of wave	$Δ x$	m	—
amplitude of wave	$a$	m	—

k_{w} = \frac{2 \cdot π}{λ_{w}}

wave number and wavelength

ω_{w} = \frac{2 \cdot π}{T_{w}}

angular frequency and period

ν_{w} = \frac{1}{T_{w}}

frequency and period

v_{w} = ν_{w} \cdot λ_{w}

speed of a wave

Δ φ_{w} = k_{w} \cdot Δ x_{w}

phase across a separation

k_{g} = \frac{2 \cdot π}{λ_{g}}

wave number and wavelength(ground)

k = \frac{2 \cdot π}{λ}

wave number and wavelength(wave)

ω_{g} = \frac{2 \cdot π}{T_{g}}

angular frequency and period(ground)

ω = \frac{2 \cdot π}{T}

angular frequency and period(wave)

ν_{g} = \frac{1}{T_{g}}

frequency and period(ground)

ν = \frac{1}{T}

frequency and period(wave)

v_{g} = ν_{g} \cdot λ_{g}

speed of a wave(ground)

v = ν \cdot λ

speed of a wave(wave)

Δ φ_{g} = k_{g} \cdot Δ x_{g}

phase across a separation(ground)

Δ φ = k \cdot Δ x

phase across a separation(wave)