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A guitar string vibrates at a frequency of 330Hz with wavelength 1.40m. The frequency and wavelength of this sound wave in air (20 C) when it reaches our ears is: Lower frequency, same wavelength Same frequency, same wavelength Higher frequency, same wavelength Same frequency, shorter wavelength

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Answer:

Same frequency, shorter wavelength

Explanation:

The speed of a wave is given by

[tex]v=f\lambda[/tex]

[tex]\lambda=\dfrac{v}{f}[/tex]

where,

f = Frequency

[tex]\lambda[/tex] = Wavelength

It can be seen that the wavelength is directly proportional to the velocity.

Here the frequency of the sound does not change.

But the velocity of the sound in air is slower.

Hence, the frequency remains same and the wavelength shortens.

In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.

Speed of wave:

here the speed of the wave should be provided by the following equation.

v = fλ

λ = v/f

here,

f = Frequency

λ = Wavelength

Also, the wavelength is directly proportional to the velocity.

So,  the frequency remains same and the wavelength shortens.

So based on this we can say that In the case when the frequency and wavelength of this sound wave in air (20 C) so it should be Same frequency, shorter wavelength.

learn more about frequency here: https://brainly.com/question/15853432?referrer=searchResults

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