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Sagot :
Answer:
[tex]\frac{T}{T_o} = ( 1 + \frac{1}{n} )^2[/tex]
Explanation:
This is a string resonance exercise, the wavelengths in a string held at the ends is
λ = 2L₀ / n
where n is an integer
the speed of the wave is
v = λ f
f = v /λ
the speed of the wave is given by the characteristics of the medium (string)
v = [tex]\sqrt{\frac{T}{\mu } }[/tex]
we substitute
f = [tex]\frac{n}{2L_o} \ \sqrt{\frac{T}{\mu } }[/tex]
to obtain the following harmonic we change n → n + 1
f’ = [tex]\frac{n+1}{2L_o} \ \sqrt{\frac{T_o}{\mu } }[/tex]
In this case, it tells us to change the tension to obtain the same frequency.
f ’= \frac{n}{2L_o} \ \sqrt{\frac{T}{\mu } }
how the two frequencies are equal
[tex]\frac{n+1}{2L_o} \sqrt{\frac{T_o}{ \mu } } = \frac{n}{2L_o} \sqrt{\frac{T}{\mu } }[/tex]
(n + 1) [tex]\sqrt{T_o}[/tex] = n [tex]\sqrt{T}[/tex]
[tex]\frac{T}{T_o} = ( \frac{n+1}{n} )^2[/tex]
[tex]\frac{T}{T_o} = ( 1 + \frac{1}{n} )^2[/tex]
this is the relationship of the voltages to obtain the following harmonic,
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