Answer
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Hint: The formula for beat frequency is the difference in frequency of the two superimposed waves.
${f_b}$ =|${f_2}−{f_1}$|
$f_1$ and $f_2$ are the frequency of two waves.
Complete step by step solution:
Given,
Length of two organ pipes, ${L_1}$ and ${L_2}$
${L_1}$= 50 cm or 0.5 m
${L_2}$= 50.5 cm or 0.505 m
Now, as we know frequency of open pipe = $\dfrac{V}{{2L}}$
Where, V is velocity of sound and L is length of pipe
$\Rightarrow {f_1}= \dfrac{V}{{2 \times 0.5}}$--------(1)
$\Rightarrow {f_2}= \dfrac{V}{{2 \times 0.505}}$-------(2)
And now beat= difference in frequency
$\Rightarrow Beat = {f_1}-{f_2}$ = $\dfrac{V}{{2 \times 0.5}}-\dfrac{{\rm{V}}}{{2 \times 0.505}}$
Given beat = 0.3
$ \Rightarrow 0.3 = \dfrac{V}{{2 \times 0.5}}-\dfrac{V}{{2 \times 0.505}}$
$\therefore V=30.3 m/sec$
Additional Information: Beat is a term used for sound waves. Beat frequency is the difference in frequency of two waves. This is due to constructive and destructive interference. In sound, we said that the beat frequency is heard as the rate at which the loudness of the sound varies while we hear the simple frequency of the waves as the pitch of the sound.
Beat frequency refers to the rate at which the volume is heard oscillating from high to low. For example, if two full cycles of high and low volume are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in the frequency of the two notes which interferes to produce the beat.
So if two sound waves with frequencies of 256 Hz and 254 Hz are run simultaneously, a beat frequency of 2 Hz will be detected.
Note: If the beat frequency is less than about 10 Hz, the two waves are very close in frequency and you will only hear one pitch (which is actually the average of the two frequencies). However, the thrust frequency will vary with frequency according to the beat frequency. Beat frequency in the range of about 10–60 Hz.
${f_b}$ =|${f_2}−{f_1}$|
$f_1$ and $f_2$ are the frequency of two waves.
Complete step by step solution:
Given,
Length of two organ pipes, ${L_1}$ and ${L_2}$
${L_1}$= 50 cm or 0.5 m
${L_2}$= 50.5 cm or 0.505 m
Now, as we know frequency of open pipe = $\dfrac{V}{{2L}}$
Where, V is velocity of sound and L is length of pipe
$\Rightarrow {f_1}= \dfrac{V}{{2 \times 0.5}}$--------(1)
$\Rightarrow {f_2}= \dfrac{V}{{2 \times 0.505}}$-------(2)
And now beat= difference in frequency
$\Rightarrow Beat = {f_1}-{f_2}$ = $\dfrac{V}{{2 \times 0.5}}-\dfrac{{\rm{V}}}{{2 \times 0.505}}$
Given beat = 0.3
$ \Rightarrow 0.3 = \dfrac{V}{{2 \times 0.5}}-\dfrac{V}{{2 \times 0.505}}$
$\therefore V=30.3 m/sec$
Additional Information: Beat is a term used for sound waves. Beat frequency is the difference in frequency of two waves. This is due to constructive and destructive interference. In sound, we said that the beat frequency is heard as the rate at which the loudness of the sound varies while we hear the simple frequency of the waves as the pitch of the sound.
Beat frequency refers to the rate at which the volume is heard oscillating from high to low. For example, if two full cycles of high and low volume are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in the frequency of the two notes which interferes to produce the beat.
So if two sound waves with frequencies of 256 Hz and 254 Hz are run simultaneously, a beat frequency of 2 Hz will be detected.
Note: If the beat frequency is less than about 10 Hz, the two waves are very close in frequency and you will only hear one pitch (which is actually the average of the two frequencies). However, the thrust frequency will vary with frequency according to the beat frequency. Beat frequency in the range of about 10–60 Hz.
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