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3 years ago

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. When two speakers play different sounds with a constant frequency, beats are heard with a frequency of 5 Hz. If the frequency

of sound played by one of the speakers is 1050 Hz, what is the frequency of sound played by the other speaker

1 answer:

morpeh [17]3 years ago

3 0

The beat frequency is the concept required to develop this process. The phenomenon is generated when you have two waves but their frequencies do not differ greatly. So the beat frequency will be the difference between those two waves. For this case we have the difference and one of the waves, therefore,

$F_b = |F_1-F_2|$

Our values are given as,

$F_b = 5Hz$

$F_1 = 1050Hz$

Using the formula of the frequency I will have,

$F_b = |F_1-F_2|$

Replacing we have,

$5 = |1050-F_2|$

$F_2 = 1050 \pm 5$

$F_2 = 1045,1055 Hz$

The possible values are two: 1045Hz and 1055Hz

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3 years ago

A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

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Answer:

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The velocity of a wave on a rope is:

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Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

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Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

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