Answer
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Hint : The velocity of sound is the same irrespective of frequency. Sounds waves with a higher frequency have a lower wavelength. So from the given values of the frequencies, we will get the velocity of sound in air.
Formula used: In this solution we will be using the following formula;
$ \Rightarrow v = f\lambda $ where $ v $ is the velocity of sound wave, $ f $ is the frequency of the wave, and $ \lambda $ is the wavelength of sound wave.
Complete step by step answer
Generally, the velocity of sound is mostly dependent on temperature, and independent on the frequency. Hence, the two sound waves have the same velocity.
Thus, for the 68 Hz frequency wave, we have $ v = {f_1}{\lambda _1} = 68{\lambda _1} $ where $ {\lambda _1} $ is the corresponding wavelength of the wave.
Similarly for the 85 Hz frequency wave, it is $ v = {f_2}{\lambda _2} = 85{\lambda _2} $
Then, we can write that $ {f_2}{\lambda _2} = {f_1}{\lambda _1} $
It is also given that, the difference between the difference in their wavelength is one metre, hence since the wavelength is higher for a lower frequency, then $ {\lambda _1} > {\lambda _2} $ , implying that
$ \Rightarrow {\lambda _1} - {\lambda _2} = 1 $
$ \Rightarrow {\lambda _1} = 1 + {\lambda _2} $
We can insert this into, the equation $ {f_2}{\lambda _2} = {f_1}{\lambda _1} $ which gives
$ \Rightarrow {f_2}{\lambda _2} = {f_1}\left( {{\lambda _2} + 1} \right) $
Hence, by inserting known values and opening the bracket, we get that
$ \Rightarrow 85{\lambda _2} = 68{\lambda _2} + 68 $
By making the wavelength the subject of the formula and can computing, we have
$ \Rightarrow {\lambda _2} = \dfrac{{65}}{{85 - 68}} = \dfrac{{68}}{{17}} = 4m $
Hence
$ \therefore v = 85 \times 4 = 340m/s $
Hence, the correct option is C.
Note
Here although, we assumed completely that the speed of sound waves is completely independent of temperature. In reality, the speed does depend on frequency however, weakly and in most cases it can be neglected.
Formula used: In this solution we will be using the following formula;
$ \Rightarrow v = f\lambda $ where $ v $ is the velocity of sound wave, $ f $ is the frequency of the wave, and $ \lambda $ is the wavelength of sound wave.
Complete step by step answer
Generally, the velocity of sound is mostly dependent on temperature, and independent on the frequency. Hence, the two sound waves have the same velocity.
Thus, for the 68 Hz frequency wave, we have $ v = {f_1}{\lambda _1} = 68{\lambda _1} $ where $ {\lambda _1} $ is the corresponding wavelength of the wave.
Similarly for the 85 Hz frequency wave, it is $ v = {f_2}{\lambda _2} = 85{\lambda _2} $
Then, we can write that $ {f_2}{\lambda _2} = {f_1}{\lambda _1} $
It is also given that, the difference between the difference in their wavelength is one metre, hence since the wavelength is higher for a lower frequency, then $ {\lambda _1} > {\lambda _2} $ , implying that
$ \Rightarrow {\lambda _1} - {\lambda _2} = 1 $
$ \Rightarrow {\lambda _1} = 1 + {\lambda _2} $
We can insert this into, the equation $ {f_2}{\lambda _2} = {f_1}{\lambda _1} $ which gives
$ \Rightarrow {f_2}{\lambda _2} = {f_1}\left( {{\lambda _2} + 1} \right) $
Hence, by inserting known values and opening the bracket, we get that
$ \Rightarrow 85{\lambda _2} = 68{\lambda _2} + 68 $
By making the wavelength the subject of the formula and can computing, we have
$ \Rightarrow {\lambda _2} = \dfrac{{65}}{{85 - 68}} = \dfrac{{68}}{{17}} = 4m $
Hence
$ \therefore v = 85 \times 4 = 340m/s $
Hence, the correct option is C.
Note
Here although, we assumed completely that the speed of sound waves is completely independent of temperature. In reality, the speed does depend on frequency however, weakly and in most cases it can be neglected.
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