
You are listening to an $A$ note played on a violin string. Let the subscript $s$ refers to the violin string and $a$ refer to the air. Then:
(A) ${f_s} = {f_a}$ but ${\lambda _s} \ne {\lambda _a}$
(B) ${f_s} = {f_a}$ but ${\lambda _s} = {\lambda _a}$
(C) ${\lambda _s} = {\lambda _a}$ but ${f_s} \ne {f_a}$
(D) ${\lambda _s} \ne {\lambda _a}$ but ${f_s} \ne {f_a}$
Answer
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Hint When the violin is played, the music is produced by the vibration of the string and the vibration of the air which are surrounding the vibrating string. And this vibration of the string and the vibration of the air lead to the formation of the frequency of the air and the frequency of the string. If there is frequency, then there is also wave length.
Complete step by step solution
Given that,
You are listening to an $A$ note played on a violin string. Let the subscript $s$ refers to the violin string and $a$ refer to the air. The sound of the violin or the music of the violin is produced by the vibration of the string and the vibration of the air. The vibration is produced by disturbing the string which is tied tightly and stretched as long as possible, then we touch the string slightly, then the vibration of the string is more. Then the violin player plays the note $A$, there are different types of sounds produced by the violin, this is due to the vibration of the string and the vibration of the air around the string.
The frequency is defined as the number of cycles per second or the number of vibrations per second. If there is a vibration, so that there will be the frequency, if there is the frequency definitely there must be the wavelength. The frequency of the string is equal to the frequency of the air because the source to produce the frequency is the same. But the air and the string are different mediums so the velocity of the sound will change, so the wavelengths are not equal.
Hence, the option (A) is the correct answer.
Note
The frequency of the sound is directly proportional to the velocity of the sound, as the velocity increases the frequency also increases. The frequency of the sound is inversely proportional to the wavelength of the sound, as the wavelength increases the frequency will decrease. $f = \dfrac{v}{\lambda }$, where, $f$ is the frequency, $v$ is the velocity and $\lambda $ is the wavelength.
Complete step by step solution
Given that,
You are listening to an $A$ note played on a violin string. Let the subscript $s$ refers to the violin string and $a$ refer to the air. The sound of the violin or the music of the violin is produced by the vibration of the string and the vibration of the air. The vibration is produced by disturbing the string which is tied tightly and stretched as long as possible, then we touch the string slightly, then the vibration of the string is more. Then the violin player plays the note $A$, there are different types of sounds produced by the violin, this is due to the vibration of the string and the vibration of the air around the string.
The frequency is defined as the number of cycles per second or the number of vibrations per second. If there is a vibration, so that there will be the frequency, if there is the frequency definitely there must be the wavelength. The frequency of the string is equal to the frequency of the air because the source to produce the frequency is the same. But the air and the string are different mediums so the velocity of the sound will change, so the wavelengths are not equal.
Hence, the option (A) is the correct answer.
Note
The frequency of the sound is directly proportional to the velocity of the sound, as the velocity increases the frequency also increases. The frequency of the sound is inversely proportional to the wavelength of the sound, as the wavelength increases the frequency will decrease. $f = \dfrac{v}{\lambda }$, where, $f$ is the frequency, $v$ is the velocity and $\lambda $ is the wavelength.
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