
A source of unknown frequency gives 4 beats/s, when sounded with a source of known frequency 250 Hz. The second harmonic of the source of unknown frequency gives five beats per seconds, when sounded with a source of frequency 513 Hz. The unknown frequency is :
A. 254 Hz
B. 246 Hz
C. 240 Hz
D. 260 Hz
Answer
486.9k+ views
Hint: Beats are produced as the result of overlapping waves. In the question the beat produced by the unknown frequency in two different cases is given. That would give us a set of possible values of the unknown frequency. The common possible value of frequency will be the frequency of the unknown source.
Formula used: \[{{f}_{uk}}=\dfrac{f}{2}\]
where \[{{f}_{uk}}\] is the unknown frequency.
Complete step by step answer:
The known frequency is given to be 250 Hz and it produces 4 beats with a unknown frequency
\[{{f}_{uk}}=250Hz\] and \[x=4beats\]
\[\Rightarrow {{f}_{uk}}=250\pm 4\] where \[{{f}_{uk}}\] is the unknown frequency.
The two possible values of \[{{f}_{uk}}\] are
\[{{f}_{uk}}=246\] or \[{{f}_{uk}}=254\]……(1)
It is also given that the unknown frequency gives 5 beats per second with a source of frequency 513 Hz
The second harmonic of unknown frequency ‘f’ will be
\[f=513\pm 5\]
\[\Rightarrow \]\[f=518\] or \[f=508\]
The correspond frequency to these second harmonics frequency will be
\[{{f}_{uk}}=\dfrac{f}{2}\]
\[{{f}_{uk}}=259\] or \[{{f}_{uk}}=254\]…..(2)
Form (1) and (2) the common possible value of unknown frequency is 254Hz, \[{{f}_{uk}}=254\]
So, the correct answer is “Option A”.
Note: Beats are produced due to a phenomenon called wave interference. These wave interferences are of two types constructive and destructive. When the same parts of the waves, i.e. crest or trough overlap then a constructive interference takes place. Meanwhile if the waves in opposite phases then destructive interference takes place.
Formula used: \[{{f}_{uk}}=\dfrac{f}{2}\]
where \[{{f}_{uk}}\] is the unknown frequency.
Complete step by step answer:
The known frequency is given to be 250 Hz and it produces 4 beats with a unknown frequency
\[{{f}_{uk}}=250Hz\] and \[x=4beats\]
\[\Rightarrow {{f}_{uk}}=250\pm 4\] where \[{{f}_{uk}}\] is the unknown frequency.
The two possible values of \[{{f}_{uk}}\] are
\[{{f}_{uk}}=246\] or \[{{f}_{uk}}=254\]……(1)
It is also given that the unknown frequency gives 5 beats per second with a source of frequency 513 Hz
The second harmonic of unknown frequency ‘f’ will be
\[f=513\pm 5\]
\[\Rightarrow \]\[f=518\] or \[f=508\]
The correspond frequency to these second harmonics frequency will be
\[{{f}_{uk}}=\dfrac{f}{2}\]
\[{{f}_{uk}}=259\] or \[{{f}_{uk}}=254\]…..(2)
Form (1) and (2) the common possible value of unknown frequency is 254Hz, \[{{f}_{uk}}=254\]
So, the correct answer is “Option A”.
Note: Beats are produced due to a phenomenon called wave interference. These wave interferences are of two types constructive and destructive. When the same parts of the waves, i.e. crest or trough overlap then a constructive interference takes place. Meanwhile if the waves in opposite phases then destructive interference takes place.
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