(C) Let $n$ be the frequency of tuning fork $C$ in $Hz$.Given that the frequency of $A$ is $3\%$ more than $C$,so $n_A = n + 0.03n = 1.03n$.The freque

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(C) Let $n$ be the frequency of tuning fork $C$ in $Hz$.Given that the frequency of $A$ is $3\%$ more than $C$,so $n_A = n + 0.03n = 1.03n$.The frequency of $B$ is $2\%$ less than $C$,so $n_B = n - 0.02n = 0.98n$.When $A$ and $B$ are sounded together,the beat frequency is given by $|n_A - n_B| = 5$.Substituting the expressions: $|1.03n - 0.98n| = 5$.$0.05n = 5$.$n = \frac{5}{0.05} = 100\,Hz$.Therefore,the frequency of tuning fork $A$ is $n_A = 1.03 \times 100 = 103\,Hz$.

Class 11 Physics Waves and Sound Beats and Tuning fork

Medium

The frequencies of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ less than the frequency of tuning fork $C$. When $A$ and $B$ are simultaneously excited,$5$ beats per second are produced. Then the frequency of the tuning fork $A$ (in $Hz$) is

A
$98$
B
$100$
C
$103$
D
$105$

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Class 11 Questions

Class 11

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Waves and Sound

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Beats and Tuning fork

The beats are produced by two sound sources of same amplitude and of nearly equal frequencies. The maximum intensity of beats will be ...... that of one source.

Easy

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The string of a violin emits a note of $205 \,Hz$ at its correct tension. The string is tightened slightly and then it produces six beats in two seconds with a tuning fork of frequency $205 \,Hz$. The frequency of the note emitted by the taut string is .......... $Hz$.

Easy

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The frequency of tuning fork $A$ is $256\,Hz$. It produces $4\,beats/sec$ with tuning fork $B$. When wax is applied to tuning fork $B$,$6\,beats/sec$ are heard. By reducing a small amount of wax,$4\,beats/sec$ are heard. The frequency of $B$ is .... $Hz$.

Medium

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The frequencies of two sound sources are $256 \text{ Hz}$ and $260 \text{ Hz}$. At $t = 0$,the intensity of sound is maximum. Then the phase difference at the time $t = \frac{1}{16} \text{ s}$ will be

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$A$ and $B$ are two wires whose fundamental frequencies are $256 \text{ Hz}$ and $382 \text{ Hz}$ respectively. When the third harmonic of $A$ and the second harmonic of $B$ are sounded together, the number of beats heard in two seconds will be:

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The frequencies of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ less than the frequency of tuning fork $C$. When $A$ and $B$ are simultaneously excited,$5$ beats per second are produced. Then the frequency of the tuning fork $A$ (in $Hz$) is

Similar Questions

The beats are produced by two sound sources of same amplitude and of nearly equal frequencies. The maximum intensity of beats will be ...... that of one source.

The string of a violin emits a note of $205 \,Hz$ at its correct tension. The string is tightened slightly and then it produces six beats in two seconds with a tuning fork of frequency $205 \,Hz$. The frequency of the note emitted by the taut string is .......... $Hz$.

The frequency of tuning fork $A$ is $256\,Hz$. It produces $4\,beats/sec$ with tuning fork $B$. When wax is applied to tuning fork $B$,$6\,beats/sec$ are heard. By reducing a small amount of wax,$4\,beats/sec$ are heard. The frequency of $B$ is .... $Hz$.

The frequencies of two sound sources are $256 \text{ Hz}$ and $260 \text{ Hz}$. At $t = 0$,the intensity of sound is maximum. Then the phase difference at the time $t = \frac{1}{16} \text{ s}$ will be

$A$ and $B$ are two wires whose fundamental frequencies are $256 \text{ Hz}$ and $382 \text{ Hz}$ respectively. When the third harmonic of $A$ and the second harmonic of $B$ are sounded together, the number of beats heard in two seconds will be:

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