Answer
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Hint: The beat can be mathematically defined as the difference between two closed frequencies. It is always defined as a positive number.
Formula used: In this solution we will be using the following formulae;
\[\Delta f = {f_1} - {f_2}\] for \[{f_1} > {f_2}\] where \[\Delta f\] is the beat of two waves with close frequencies, \[{f_1}\] is the frequency of one of the wave and \[{f_2}\] is the frequency of the other.
Complete Step-by-Step solution:
Generally, a beat can be defined as an interference between two waves which are only slightly different in frequency. The resulting wave is usually a periodic increase and decrease of the amplitude which in itself has a frequency which is the difference between the two initial waves.
Hence, mathematically, the beat frequency is given as
\[\Delta f = {f_1} - {f_2}\] for \[{f_1} > {f_2}\] where \[\Delta f\] is the beat of two waves with close frequencies, \[{f_1}\] is the frequency of one of the wave and \[{f_2}\] is the frequency of the other.
We see that it is always a positive number. In above, if \[{f_2} > {f_1}\] the equation would be
\[\Delta f = {f_2} - {f_1}\]
In the question, we are told that the beat between the oscillator and tuning fork is 2 Hz when the oscillator produces 514 Hz. So, the equation may be
\[\Delta f = {f_o} - {f_t} \Rightarrow 2 = 514 - {f_t}\]
\[{f_t} = 512Hz\]
where “o” and “t” signifies an oscillator and tuning fork respectively.
It may also be
\[\Delta f = {f_t} - {f_o} \Rightarrow 2 = {f_t} - 514\]
\[{f_t} = 516Hz\]
Hence, we need a second equation to be certain. We are told that for 510 Hz of oscillator we have a beat of 6 Hz.
So, if it were 512 Hz, the beat in the second situation would have been 2 Hz, and not 6 Hz.
Hence, the correct option is 516 Hz.
Thus, the correct option is A.
Note: Alternatively, for examination purposes, we can observe the options for clues. If we look at the option, we see that there are no 512 Hz as the answer. Hence, even using the first equation alone, we can conclude that the correct answer is 516 Hz.
Formula used: In this solution we will be using the following formulae;
\[\Delta f = {f_1} - {f_2}\] for \[{f_1} > {f_2}\] where \[\Delta f\] is the beat of two waves with close frequencies, \[{f_1}\] is the frequency of one of the wave and \[{f_2}\] is the frequency of the other.
Complete Step-by-Step solution:
Generally, a beat can be defined as an interference between two waves which are only slightly different in frequency. The resulting wave is usually a periodic increase and decrease of the amplitude which in itself has a frequency which is the difference between the two initial waves.
Hence, mathematically, the beat frequency is given as
\[\Delta f = {f_1} - {f_2}\] for \[{f_1} > {f_2}\] where \[\Delta f\] is the beat of two waves with close frequencies, \[{f_1}\] is the frequency of one of the wave and \[{f_2}\] is the frequency of the other.
We see that it is always a positive number. In above, if \[{f_2} > {f_1}\] the equation would be
\[\Delta f = {f_2} - {f_1}\]
In the question, we are told that the beat between the oscillator and tuning fork is 2 Hz when the oscillator produces 514 Hz. So, the equation may be
\[\Delta f = {f_o} - {f_t} \Rightarrow 2 = 514 - {f_t}\]
\[{f_t} = 512Hz\]
where “o” and “t” signifies an oscillator and tuning fork respectively.
It may also be
\[\Delta f = {f_t} - {f_o} \Rightarrow 2 = {f_t} - 514\]
\[{f_t} = 516Hz\]
Hence, we need a second equation to be certain. We are told that for 510 Hz of oscillator we have a beat of 6 Hz.
So, if it were 512 Hz, the beat in the second situation would have been 2 Hz, and not 6 Hz.
Hence, the correct option is 516 Hz.
Thus, the correct option is A.
Note: Alternatively, for examination purposes, we can observe the options for clues. If we look at the option, we see that there are no 512 Hz as the answer. Hence, even using the first equation alone, we can conclude that the correct answer is 516 Hz.
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