Two Sitar Strings

solved Beats

Q. Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz $ν_{beat,ab} = 5$ Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz $ν_{beat,at} = 3$ Hz. What is the original frequency of B if the frequency of A is 427 Hz $ν_{a} = 427$ Hz, in hertz?

Inputs

beats

ν_{beat,ab}

5 Hz

fewer beats

ν_{beat,at}

3 Hz

frequency a

ν_{a}

427 Hz

Outputs

frequency

ν_{b}

How it's solved

Solved with Beats

Given

$ν_{a} = 427$ Hz
$ν_{beat,ab} = 5$ Hz
$ν_{beat,at} = 3$ Hz

Find $ν_{b}$ .

Working

beats of two sounds(a, b) $ν_{beat,ab} = ν_{a} - ν_{b}$
$5 = 427 - ν_{b}$
$ν_{b} = 422$ Hz

Conclusion

Tightening B raises its frequency; since the beats then slow down, B was moving towards A from below, so A beats against B from above.

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 :two_sitar_strings do4  kind :numerical5  describes "Two sitar strings A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency " \6            "%{beats}. The tension of the string B is slightly increased and the beat frequency is found to decrease " \7            "to %{fewer_beats}. %{frequency} if the frequency of A is %{frequency_a}, in hertz?"8  given beats: "5 Hz"9  given fewer_beats: "3 Hz"10  given frequency_a: "427 Hz"11  asks frequency: "What is the original frequency of B"12end

solution.rb

1require_relative "../questions/two_sitar_strings"2require_relative "../science/waves/beats"34solve :two_sitar_strings, using: Beats do5  objects :a, :b, :tightened_b67  take frequency_a: a.frequency8  take beats: a.beat_frequency(by: b)9  take fewer_beats: a.beat_frequency(by: tightened_b)1011  frequency = b.frequency12  answer frequency:, in: "Hz"1314  conclusion "Tightening B raises its frequency; since the beats then slow down, B was moving towards A " \15             "from below, so A beats against B from above."16end

playground.rb

1require_relative "../solutions/two_sitar_strings"23play_with :two_sitar_strings do4  title "Two Sitar Strings"5end

The laws it drew on

quantity	written	units	within	from
frequency	$ν$	Hz	0 … ∞	Beats
beat frequency	$ν_{beat}$	Hz	—	Beats
beat frequency of ground by a	$ν_{beat,ga}$	Hz	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
frequency of a	$ν_{a}$	Hz	0 … ∞
beat frequency of ground by b	$ν_{beat,gb}$	Hz	—
frequency of b	$ν_{b}$	Hz	0 … ∞
beat frequency of ground by tightened b	$ν_{beat,gt}$	Hz	—
frequency of tightened b	$ν_{t}$	Hz	0 … ∞
beat frequency of a by ground	$ν_{beat,ag}$	Hz	—
beat frequency of a by b	$ν_{beat,ab}$	Hz	—
beat frequency of a by tightened b	$ν_{beat,at}$	Hz	—
beat frequency of b by ground	$ν_{beat,bg}$	Hz	—
beat frequency of b by a	$ν_{beat,ba}$	Hz	—
beat frequency of b by tightened b	$ν_{beat,bt}$	Hz	—
beat frequency of tightened b by ground	$ν_{beat,tg}$	Hz	—
beat frequency of tightened b by a	$ν_{beat,ta}$	Hz	—
beat frequency of tightened b by b	$ν_{beat,tb}$	Hz	—

ν_{beat,oo} = ν_{o} - ν_{o}

beats of two sounds

ν_{beat,ga} = ν_{g} - ν_{a}

beats of two sounds(ground, a)

ν_{beat,gb} = ν_{g} - ν_{b}

beats of two sounds(ground, b)

ν_{beat,gt} = ν_{g} - ν_{t}

beats of two sounds(ground, tightened b)

ν_{beat,ag} = ν_{a} - ν_{g}

beats of two sounds(a, ground)

ν_{beat,ab} = ν_{a} - ν_{b}

beats of two sounds(a, b)

ν_{beat,at} = ν_{a} - ν_{t}

beats of two sounds(a, tightened b)

ν_{beat,bg} = ν_{b} - ν_{g}

beats of two sounds(b, ground)

ν_{beat,ba} = ν_{b} - ν_{a}

beats of two sounds(b, a)

ν_{beat,bt} = ν_{b} - ν_{t}

beats of two sounds(b, tightened b)

ν_{beat,tg} = ν_{t} - ν_{g}

beats of two sounds(tightened b, ground)

ν_{beat,ta} = ν_{t} - ν_{a}

beats of two sounds(tightened b, a)

ν_{beat,tb} = ν_{t} - ν_{b}

beats of two sounds(tightened b, b)