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PHXI15:WAVES

354982 When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 H z$ . When one of the prongs of the fork $A$ is filed and sounded with $B$ , the beat frequency increases, then the frequency of the fork $A$ is

1

388 H z

2

389 H z

3

380 H z

4

379 H z

Explanation:

Let the frequency of the fork $A$ be $f_{A} .$
As it produces 4 beats per second with the fork $B$ of frequency $f_{B} = 384 H z$ ,
$∴ f_{A} = f_{B} \pm 4 = (384 \pm 4) H z$
$= 388 H z$ or $380 H z$
When one of the prongs of $A$ is filed, its frequency increases.
If $f_{A} = 380 H z$ , further increase in $f_{A}$ will result in decrease in the beat frequency when sounded with $B$ .
If $f_{A} = 388 H z$ , further increase in $f_{A}$ will result in increase in the beat frequency when sounded with $B$ .
Thus, the frequency of the fork $A$ is $388 H z$ .

PHXI15:WAVES

354983 A closed and an open organ pipe of same length are set into vibrations simultaneously in their fundamental mode to produce 2 beats. The length of open organ pipe is now halved and of closed organ pipe is doubled. Now find the number of beats produced.

1 7

2 9

3 5

4 10

Explanation:

$\begin{aligned} f_{0} - f_{c} = 2 \\ V [\frac{1}{2 L} - \frac{1}{4 L}] = 2 or V / L = 8 \end{aligned}$
In the second case,
$f_{0}^{'} - f_{c}^{'} = \frac{V}{L} - \frac{V}{8 L} = \frac{7 V}{8 L} = \frac{7}{8} (8) = 7$

PHXI15:WAVES

354984 When beats are formed by two waves of frequencies $n_{1}$ and $n_{2}$ , the amplitude varies with frequency equal to

1

n_{1} - n_{2}

2

2 (n_{1} - n_{2})

3

(n_{1} - n_{2}) / 2

4

(n_{1} + n_{2}) / 2

PHXI15:WAVES

354985 25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first tuning fork is the octave of the last fork, then the frequency of the $21^{s t}$ fork is

1

288 H z

2

72 H z

3

87 H z

4

84 H z

Explanation:

According to the question frequencies of first and last tuning forks are $2 n$ and $n$ respectively.
Hence frequency in given arrangement are as follows
$n_{Last} = n_{first} + (N - 1) x$
$\Rightarrow 2 n - 24 \times 3 = n \Rightarrow n = 72 H z$
So, frequency of $21^{st}$ tuning fork
$n_{21} = (2 \times 72 - 20 \times 3) = 84 H z .$

PHXI15:WAVES

354982 When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 H z$ . When one of the prongs of the fork $A$ is filed and sounded with $B$ , the beat frequency increases, then the frequency of the fork $A$ is

1

388 H z

2

389 H z

3

380 H z

4

379 H z

Explanation:

Let the frequency of the fork $A$ be $f_{A} .$
As it produces 4 beats per second with the fork $B$ of frequency $f_{B} = 384 H z$ ,
$∴ f_{A} = f_{B} \pm 4 = (384 \pm 4) H z$
$= 388 H z$ or $380 H z$
When one of the prongs of $A$ is filed, its frequency increases.
If $f_{A} = 380 H z$ , further increase in $f_{A}$ will result in decrease in the beat frequency when sounded with $B$ .
If $f_{A} = 388 H z$ , further increase in $f_{A}$ will result in increase in the beat frequency when sounded with $B$ .
Thus, the frequency of the fork $A$ is $388 H z$ .

PHXI15:WAVES

354983 A closed and an open organ pipe of same length are set into vibrations simultaneously in their fundamental mode to produce 2 beats. The length of open organ pipe is now halved and of closed organ pipe is doubled. Now find the number of beats produced.

1 7

2 9

3 5

4 10

Explanation:

$\begin{aligned} f_{0} - f_{c} = 2 \\ V [\frac{1}{2 L} - \frac{1}{4 L}] = 2 or V / L = 8 \end{aligned}$
In the second case,
$f_{0}^{'} - f_{c}^{'} = \frac{V}{L} - \frac{V}{8 L} = \frac{7 V}{8 L} = \frac{7}{8} (8) = 7$

PHXI15:WAVES

354984 When beats are formed by two waves of frequencies $n_{1}$ and $n_{2}$ , the amplitude varies with frequency equal to

1

n_{1} - n_{2}

2

2 (n_{1} - n_{2})

3

(n_{1} - n_{2}) / 2

4

(n_{1} + n_{2}) / 2

PHXI15:WAVES

354985 25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first tuning fork is the octave of the last fork, then the frequency of the $21^{s t}$ fork is

1

288 H z

2

72 H z

3

87 H z

4

84 H z

Explanation:

According to the question frequencies of first and last tuning forks are $2 n$ and $n$ respectively.
Hence frequency in given arrangement are as follows
$n_{Last} = n_{first} + (N - 1) x$
$\Rightarrow 2 n - 24 \times 3 = n \Rightarrow n = 72 H z$
So, frequency of $21^{st}$ tuning fork
$n_{21} = (2 \times 72 - 20 \times 3) = 84 H z .$

PHXI15:WAVES

354982 When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 H z$ . When one of the prongs of the fork $A$ is filed and sounded with $B$ , the beat frequency increases, then the frequency of the fork $A$ is

1

388 H z

2

389 H z

3

380 H z

4

379 H z

Explanation:

Let the frequency of the fork $A$ be $f_{A} .$
As it produces 4 beats per second with the fork $B$ of frequency $f_{B} = 384 H z$ ,
$∴ f_{A} = f_{B} \pm 4 = (384 \pm 4) H z$
$= 388 H z$ or $380 H z$
When one of the prongs of $A$ is filed, its frequency increases.
If $f_{A} = 380 H z$ , further increase in $f_{A}$ will result in decrease in the beat frequency when sounded with $B$ .
If $f_{A} = 388 H z$ , further increase in $f_{A}$ will result in increase in the beat frequency when sounded with $B$ .
Thus, the frequency of the fork $A$ is $388 H z$ .

PHXI15:WAVES

354983 A closed and an open organ pipe of same length are set into vibrations simultaneously in their fundamental mode to produce 2 beats. The length of open organ pipe is now halved and of closed organ pipe is doubled. Now find the number of beats produced.

1 7

2 9

3 5

4 10

Explanation:

$\begin{aligned} f_{0} - f_{c} = 2 \\ V [\frac{1}{2 L} - \frac{1}{4 L}] = 2 or V / L = 8 \end{aligned}$
In the second case,
$f_{0}^{'} - f_{c}^{'} = \frac{V}{L} - \frac{V}{8 L} = \frac{7 V}{8 L} = \frac{7}{8} (8) = 7$

PHXI15:WAVES

354984 When beats are formed by two waves of frequencies $n_{1}$ and $n_{2}$ , the amplitude varies with frequency equal to

1

n_{1} - n_{2}

2

2 (n_{1} - n_{2})

3

(n_{1} - n_{2}) / 2

4

(n_{1} + n_{2}) / 2

PHXI15:WAVES

354985 25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first tuning fork is the octave of the last fork, then the frequency of the $21^{s t}$ fork is

1

288 H z

2

72 H z

3

87 H z

4

84 H z

Explanation:

According to the question frequencies of first and last tuning forks are $2 n$ and $n$ respectively.
Hence frequency in given arrangement are as follows
$n_{Last} = n_{first} + (N - 1) x$
$\Rightarrow 2 n - 24 \times 3 = n \Rightarrow n = 72 H z$
So, frequency of $21^{st}$ tuning fork
$n_{21} = (2 \times 72 - 20 \times 3) = 84 H z .$

PHXI15:WAVES

354982 When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384 H z$ . When one of the prongs of the fork $A$ is filed and sounded with $B$ , the beat frequency increases, then the frequency of the fork $A$ is

1

388 H z

2

389 H z

3

380 H z

4

379 H z

Explanation:

Let the frequency of the fork $A$ be $f_{A} .$
As it produces 4 beats per second with the fork $B$ of frequency $f_{B} = 384 H z$ ,
$∴ f_{A} = f_{B} \pm 4 = (384 \pm 4) H z$
$= 388 H z$ or $380 H z$
When one of the prongs of $A$ is filed, its frequency increases.
If $f_{A} = 380 H z$ , further increase in $f_{A}$ will result in decrease in the beat frequency when sounded with $B$ .
If $f_{A} = 388 H z$ , further increase in $f_{A}$ will result in increase in the beat frequency when sounded with $B$ .
Thus, the frequency of the fork $A$ is $388 H z$ .

PHXI15:WAVES

354983 A closed and an open organ pipe of same length are set into vibrations simultaneously in their fundamental mode to produce 2 beats. The length of open organ pipe is now halved and of closed organ pipe is doubled. Now find the number of beats produced.

1 7

2 9

3 5

4 10

Explanation:

$\begin{aligned} f_{0} - f_{c} = 2 \\ V [\frac{1}{2 L} - \frac{1}{4 L}] = 2 or V / L = 8 \end{aligned}$
In the second case,
$f_{0}^{'} - f_{c}^{'} = \frac{V}{L} - \frac{V}{8 L} = \frac{7 V}{8 L} = \frac{7}{8} (8) = 7$

PHXI15:WAVES

354984 When beats are formed by two waves of frequencies $n_{1}$ and $n_{2}$ , the amplitude varies with frequency equal to

1

n_{1} - n_{2}

2

2 (n_{1} - n_{2})

3

(n_{1} - n_{2}) / 2

4

(n_{1} + n_{2}) / 2

PHXI15:WAVES

354985 25 tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce 3 beats/sec. If the frequency of the first tuning fork is the octave of the last fork, then the frequency of the $21^{s t}$ fork is

1

288 H z

2

72 H z

3

87 H z

4

84 H z

Explanation:

According to the question frequencies of first and last tuning forks are $2 n$ and $n$ respectively.
Hence frequency in given arrangement are as follows
$n_{Last} = n_{first} + (N - 1) x$
$\Rightarrow 2 n - 24 \times 3 = n \Rightarrow n = 72 H z$
So, frequency of $21^{st}$ tuning fork
$n_{21} = (2 \times 72 - 20 \times 3) = 84 H z .$

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