Physics, 20.07.2020 21:01 shelbybibb99

Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una tension doble? Y si la tension se reduce a la mitad?

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Answer from: stephenandvicki

a) f = 615.2 Hz b) f = 307.6 Hz

Explanation:

The speed in a wave on a string is

v = √ T / μ

also the speed a wave must meet the relationship

v = λ f

Let's use these expressions in our problem, for the initial conditions

v = √ T₀ /μ

√ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

T = 2T₀

√ (T /μ) = λ f

√( 2To /μ) = λ f

√2 √ T₀ /μ = λ f

we substitute

√2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

λ₀ = λ

f = f₀ √2

f = 435 √ 2

f = 615.2 Hz

b) The tension is cut in half

T = T₀ / 2

√ (T₀ / 2muy) = f = λ f

√ (T₀ / μ) 1 /√2 = λ f

fo / √2 = f

f = 435 / √2

f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

v= λ f

Usemos estas expresión en nuestro problema, para las condiciones iniciales

v= √ To/μ

√ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

T = 2T₀

√ ( T/μ) = λf

√ ) 2T₀/μ = λ f

√ 2 √ T₀/μ = λ f

substituimos

√2 ( λ₀ f₀) = λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

λ₀ = λ

f = f₀ √2

f = 435 √2

f = 615,2 Hz

b) La tension se reduce a la mitad

T = T₀/2

RA ( T₀/2μ) = λ f

Ra(T₀/μ) 1/ra 2 = λ f

fo /√ 2 = f

f = 435/√2

f = 307,6 Hz

Answer from: Quest

it should be -8.6 cm hope this out

Answer from: Quest

answer: he got 23

explanation:

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Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibr...

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