Two Sitar Strings Playing Ga

solved Beats

Q. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz $ν_{beat,ab} = 6$ Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz $ν_{beat,sb} = 3$ Hz. If the original frequency of A is 324 Hz $ν_{a} = 324$ Hz, what is the frequency of B, in hertz?

Inputs

beats

ν_{beat,ab}

6 Hz

fewer beats

ν_{beat,sb}

3 Hz

frequency a

ν_{a}

324 Hz

Outputs

frequency

ν_{b}

How it's solved

Solved with Beats

Given

$ν_{a} = 324$ Hz
$ν_{beat,ab} = 6$ Hz
$ν_{beat,sb} = 3$ Hz

Find $ν_{b}$ .

Working

beats of two sounds(a, b) $ν_{beat,ab} = ν_{a} - ν_{b}$
$6 = 324 - ν_{b}$
$ν_{b} = 318$ Hz

Conclusion

Slackening A lowers its frequency; since the beats then slow down, A was coming down towards B from above, so B lies below A.

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_playing_ga do4  kind :numerical5  describes "Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency " \6            "%{beats}. The tension in the string A is slightly reduced and the beat frequency is found to reduce to " \7            "%{fewer_beats}. If the original frequency of A is %{frequency_a}, %{frequency}, in hertz?"8  given beats: "6 Hz"9  given fewer_beats: "3 Hz"10  given frequency_a: "324 Hz"11  asks frequency: "what is the frequency of B"12end

solution.rb

1require_relative "../questions/two_sitar_strings_playing_ga"2require_relative "../science/waves/beats"34solve :two_sitar_strings_playing_ga, using: Beats do5  objects :a, :b, :slackened_a67  take frequency_a: a.frequency8  take beats: a.beat_frequency(by: b)9  take fewer_beats: slackened_a.beat_frequency(by: b)1011  frequency = b.frequency12  answer frequency:, in: "Hz"1314  conclusion "Slackening A lowers its frequency; since the beats then slow down, A was coming down towards B " \15             "from above, so B lies below A."16end

playground.rb

1require_relative "../solutions/two_sitar_strings_playing_ga"23play_with :two_sitar_strings_playing_ga do4  title "Two Sitar Strings Playing Ga"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 slackened a	$ν_{beat,gs}$	Hz	—
frequency of slackened a	$ν_{s}$	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 slackened a	$ν_{beat,as}$	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 slackened a	$ν_{beat,bs}$	Hz	—
beat frequency of slackened a by ground	$ν_{beat,sg}$	Hz	—
beat frequency of slackened a by a	$ν_{beat,sa}$	Hz	—
beat frequency of slackened a by b	$ν_{beat,sb}$	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,gs} = ν_{g} - ν_{s}

beats of two sounds(ground, slackened a)

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,as} = ν_{a} - ν_{s}

beats of two sounds(a, slackened a)

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,bs} = ν_{b} - ν_{s}

beats of two sounds(b, slackened a)

ν_{beat,sg} = ν_{s} - ν_{g}

beats of two sounds(slackened a, ground)

ν_{beat,sa} = ν_{s} - ν_{a}

beats of two sounds(slackened a, a)

ν_{beat,sb} = ν_{s} - ν_{b}

beats of two sounds(slackened a, b)