(B) The beat frequency is the absolute difference between the two frequencies.Given,for frequency $\nu$ and $200 \;Hz$,the beat frequency is $5 \;Hz$.

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(B) The beat frequency is the absolute difference between the two frequencies.Given,for frequency $\nu$ and $200 \;Hz$,the beat frequency is $5 \;Hz$.So,$\nu = 200 \pm 5$,which means $\nu = 195 \;Hz$ or $\nu = 205 \;Hz$ ... $(i)$For the second harmonic $2\nu$ and $420 \;Hz$,the beat frequency is $10 \;Hz$.So,$2\nu = 420 \pm 10$,which means $2\nu = 410 \;Hz$ or $2\nu = 430 \;Hz$.Dividing by $2$,we get $\nu = 205 \;Hz$ or $\nu = 215 \;Hz$ ... $(ii)$Comparing equations $(i)$ and $(ii)$,the common value is $\nu = 205 \;Hz$.

Class 11 Physics Waves and Sound Beats and Tuning fork

Medium

$A$ source of frequency $\nu$ gives $5$ beats/second when sounded with a source of frequency $200 \;Hz$. The second harmonic of frequency $2\nu$ of the source gives $10$ beats/second when sounded with a source of frequency $420 \;Hz$. The value of $\nu$ is .... $Hz$.

AIPMT 1994 All AIPMT

A
$195$
B
$205$
C
$200$
D
$210$

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Waves and Sound

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Beats and Tuning fork

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AIPMT 1994 Paper All AIPMT years →

Two sound waves with wavelengths $5.0\, m$ and $5.5\, m$ respectively,each propagate in a gas with velocity $330\, m/s$. We expect the following number of beats per second.

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$A$ tuning fork of frequency $280\, Hz$ produces $10$ beats per second when sounded with a vibrating sonometer string. When the tension in the string increases slightly,it produces $11$ beats per second. The original frequency of the vibrating sonometer string is ... $Hz$.

Medium

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Two identical piano wires have a fundamental frequency of $600 \text{ Hz}$ when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of $6$ beats per second when both wires vibrate simultaneously?

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$A$ string under a tension of $129.6 \ N$ produces $10 \ beats/s$ when it is vibrated along with a tuning fork. When the tension in the string is increased to $160 \ N$,it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork in $Hz$.

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$A$ source of frequency $\nu$ gives $5$ beats/second when sounded with a source of frequency $200 \;Hz$. The second harmonic of frequency $2\nu$ of the source gives $10$ beats/second when sounded with a source of frequency $420 \;Hz$. The value of $\nu$ is .... $Hz$.

Similar Questions

Two sound waves with wavelengths $5.0\, m$ and $5.5\, m$ respectively,each propagate in a gas with velocity $330\, m/s$. We expect the following number of beats per second.

Two tuning forks of frequencies $320 \,Hz$ and $480 \,Hz$ are sounded together to produce sound waves. The velocity of sound in air is $320 \,ms^{-1}$. The difference between the wavelengths of these waves is nearly: (in $cm$)

$A$ tuning fork of frequency $280\, Hz$ produces $10$ beats per second when sounded with a vibrating sonometer string. When the tension in the string increases slightly,it produces $11$ beats per second. The original frequency of the vibrating sonometer string is ... $Hz$.

Two identical piano wires have a fundamental frequency of $600 \text{ Hz}$ when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of $6$ beats per second when both wires vibrate simultaneously?

$A$ string under a tension of $129.6 \ N$ produces $10 \ beats/s$ when it is vibrated along with a tuning fork. When the tension in the string is increased to $160 \ N$,it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork in $Hz$.

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