
The frequency of a sound wave is n and its velocity is v, if frequency is increased to 4n, the velocities of the wave will be:
(A) v
(B) 4v
(C) 2v
(D) \[\dfrac{v}{4}\]
Answer
232.8k+ views
Hint In this question we are given the frequency of sound and its velocity. We have to find the change in velocity when its frequency becomes four fold. We know that the velocity of a wave is inversely proportional to its time period. This tells us that it is directly proportional to its frequency.
Complete step-by-step solution
We know that frequency of a sound wave is the reciprocal of the time period of the wave, so
\[v = \dfrac{1}{T}\]
The speed of a sound wave is expressed as
\[v = \dfrac{\lambda }{T}\]
\[v = \lambda n\] -- 1
When the frequency is increased to a value of 4n
\[{v_2} = \lambda (4n)\] -- 2
Dividing 2 by 1
\[\dfrac{{{v_2}}}{v} = \dfrac{{\lambda (4n)}}{{\lambda n}}\]
\[{v_2} = 4v\]
Therefore, the correct answer is option B
Note: Frequency means the number of distance in unit time, if frequency is increased as per the question, it means that the wave will now oscillate more in 1 second, this will ultimately have an impact on the speed of the wave if the wavelength is kept constant, i.e. it will cover more distance.
Complete step-by-step solution
We know that frequency of a sound wave is the reciprocal of the time period of the wave, so
\[v = \dfrac{1}{T}\]
The speed of a sound wave is expressed as
\[v = \dfrac{\lambda }{T}\]
\[v = \lambda n\] -- 1
When the frequency is increased to a value of 4n
\[{v_2} = \lambda (4n)\] -- 2
Dividing 2 by 1
\[\dfrac{{{v_2}}}{v} = \dfrac{{\lambda (4n)}}{{\lambda n}}\]
\[{v_2} = 4v\]
Therefore, the correct answer is option B
Note: Frequency means the number of distance in unit time, if frequency is increased as per the question, it means that the wave will now oscillate more in 1 second, this will ultimately have an impact on the speed of the wave if the wavelength is kept constant, i.e. it will cover more distance.
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