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PHXI15:WAVES

354995 A tuning fork $A$ produces 4 beats $s^{- 1}$ with another tuning fork $B$ of frequency $320 H z .$ On filling one of the prongs of $A, 4$ beats $s^{- 1}$ are again heard when sounded with the same fork $B$ . Then, the frequency of the fork $A$ before filling is

1

328 H z

2

316 H z

3

324 H z

4

320 H z

Explanation:

There are 4 beats between $A$ and $B$ , therefore the possible frequencies of $A$ are 316 or 324 , $i . e .$ $(320 \pm 4) H z$ .
When the prong of $A$ is filled, its frequency becomes greater than the original frequency. If we assume that original frequency of $A$ is 324 , then on filling its frequency will be greater than 324. The beats between $A$ and $B$ will be more than 4.
But it is given that, the beats are again 4, therefore, 324 is not possible.
Therefore, required frequency must be $316 H z .$ This is true because on filing the frequency may increase so as to give 4 beats with $B$ of frequency $320 H z .$

PHXI15:WAVES

354996 Two tuning forks, $A$ and $B$ , produce notes of frequencies $256 H z$ and $262 H z$ . An unknown note sounded with A produces certain beats. When the same note is sounded with $B$ , the beat frequency gets doubled. The unknown frequency is

1

250 H z

2

252 H z

3

254 H z

4

256 H z

Explanation:

$262 - f = 2 (256 - f)$
$f = 250 H z$

PHXI15:WAVES

354997 Two identical flutes produce fundamental notes of frequency $300 H z$ at $27^{\circ} C$ . If the temperature of air in one flute is increased to $31^{\circ} C$ , the number of the beats heard per second will be

1 2

2 1

3 4

4 3

Explanation:

Velocity of sound increases if the temperature increases. So with $v = n λ$ , if $v$ increases $n$ will increse at $27^{\circ} C, v_{1} = n λ, at 31^{\circ} C, v_{2} = (n + x) λ$
Now using $v \propto \sqrt{T} (∵ v = \sqrt{\frac{γ R T}{M}})$
$\Rightarrow \frac{300 + x}{300} = \sqrt{\frac{(273 + 31)}{(273 + 27)}} = \sqrt{\frac{304}{300}} = \sqrt{\frac{300 + 4}{300}}$
$\Rightarrow 1 + \frac{x}{300} = {(1 + \frac{4}{300})}^{1 / 2} = (1 + \frac{1}{2} \times \frac{4}{300}) \Rightarrow x = 2$ .
$[∵ (1 + x)^{n} = 1 + n x]$

PHXI15:WAVES

354998 At a point, beat frequency of $n H z$ is observed. It means:

1 Amplitude of vibrations changes simple harmonically with frequency

n / 2 H z

at that point only

2 Medium particles, at that point, are vibrating with frequency

n H z

3 At that, zero intensity is observed

2 n

times per second

4 None of the above

PHXI15:WAVES

354995 A tuning fork $A$ produces 4 beats $s^{- 1}$ with another tuning fork $B$ of frequency $320 H z .$ On filling one of the prongs of $A, 4$ beats $s^{- 1}$ are again heard when sounded with the same fork $B$ . Then, the frequency of the fork $A$ before filling is

1

328 H z

2

316 H z

3

324 H z

4

320 H z

Explanation:

There are 4 beats between $A$ and $B$ , therefore the possible frequencies of $A$ are 316 or 324 , $i . e .$ $(320 \pm 4) H z$ .
When the prong of $A$ is filled, its frequency becomes greater than the original frequency. If we assume that original frequency of $A$ is 324 , then on filling its frequency will be greater than 324. The beats between $A$ and $B$ will be more than 4.
But it is given that, the beats are again 4, therefore, 324 is not possible.
Therefore, required frequency must be $316 H z .$ This is true because on filing the frequency may increase so as to give 4 beats with $B$ of frequency $320 H z .$

PHXI15:WAVES

354996 Two tuning forks, $A$ and $B$ , produce notes of frequencies $256 H z$ and $262 H z$ . An unknown note sounded with A produces certain beats. When the same note is sounded with $B$ , the beat frequency gets doubled. The unknown frequency is

1

250 H z

2

252 H z

3

254 H z

4

256 H z

Explanation:

$262 - f = 2 (256 - f)$
$f = 250 H z$

PHXI15:WAVES

354997 Two identical flutes produce fundamental notes of frequency $300 H z$ at $27^{\circ} C$ . If the temperature of air in one flute is increased to $31^{\circ} C$ , the number of the beats heard per second will be

1 2

2 1

3 4

4 3

Explanation:

Velocity of sound increases if the temperature increases. So with $v = n λ$ , if $v$ increases $n$ will increse at $27^{\circ} C, v_{1} = n λ, at 31^{\circ} C, v_{2} = (n + x) λ$
Now using $v \propto \sqrt{T} (∵ v = \sqrt{\frac{γ R T}{M}})$
$\Rightarrow \frac{300 + x}{300} = \sqrt{\frac{(273 + 31)}{(273 + 27)}} = \sqrt{\frac{304}{300}} = \sqrt{\frac{300 + 4}{300}}$
$\Rightarrow 1 + \frac{x}{300} = {(1 + \frac{4}{300})}^{1 / 2} = (1 + \frac{1}{2} \times \frac{4}{300}) \Rightarrow x = 2$ .
$[∵ (1 + x)^{n} = 1 + n x]$

PHXI15:WAVES

354998 At a point, beat frequency of $n H z$ is observed. It means:

1 Amplitude of vibrations changes simple harmonically with frequency

n / 2 H z

at that point only

2 Medium particles, at that point, are vibrating with frequency

n H z

3 At that, zero intensity is observed

2 n

times per second

4 None of the above

PHXI15:WAVES

354995 A tuning fork $A$ produces 4 beats $s^{- 1}$ with another tuning fork $B$ of frequency $320 H z .$ On filling one of the prongs of $A, 4$ beats $s^{- 1}$ are again heard when sounded with the same fork $B$ . Then, the frequency of the fork $A$ before filling is

1

328 H z

2

316 H z

3

324 H z

4

320 H z

Explanation:

There are 4 beats between $A$ and $B$ , therefore the possible frequencies of $A$ are 316 or 324 , $i . e .$ $(320 \pm 4) H z$ .
When the prong of $A$ is filled, its frequency becomes greater than the original frequency. If we assume that original frequency of $A$ is 324 , then on filling its frequency will be greater than 324. The beats between $A$ and $B$ will be more than 4.
But it is given that, the beats are again 4, therefore, 324 is not possible.
Therefore, required frequency must be $316 H z .$ This is true because on filing the frequency may increase so as to give 4 beats with $B$ of frequency $320 H z .$

PHXI15:WAVES

354996 Two tuning forks, $A$ and $B$ , produce notes of frequencies $256 H z$ and $262 H z$ . An unknown note sounded with A produces certain beats. When the same note is sounded with $B$ , the beat frequency gets doubled. The unknown frequency is

1

250 H z

2

252 H z

3

254 H z

4

256 H z

Explanation:

$262 - f = 2 (256 - f)$
$f = 250 H z$

PHXI15:WAVES

354997 Two identical flutes produce fundamental notes of frequency $300 H z$ at $27^{\circ} C$ . If the temperature of air in one flute is increased to $31^{\circ} C$ , the number of the beats heard per second will be

1 2

2 1

3 4

4 3

Explanation:

Velocity of sound increases if the temperature increases. So with $v = n λ$ , if $v$ increases $n$ will increse at $27^{\circ} C, v_{1} = n λ, at 31^{\circ} C, v_{2} = (n + x) λ$
Now using $v \propto \sqrt{T} (∵ v = \sqrt{\frac{γ R T}{M}})$
$\Rightarrow \frac{300 + x}{300} = \sqrt{\frac{(273 + 31)}{(273 + 27)}} = \sqrt{\frac{304}{300}} = \sqrt{\frac{300 + 4}{300}}$
$\Rightarrow 1 + \frac{x}{300} = {(1 + \frac{4}{300})}^{1 / 2} = (1 + \frac{1}{2} \times \frac{4}{300}) \Rightarrow x = 2$ .
$[∵ (1 + x)^{n} = 1 + n x]$

PHXI15:WAVES

354998 At a point, beat frequency of $n H z$ is observed. It means:

1 Amplitude of vibrations changes simple harmonically with frequency

n / 2 H z

at that point only

2 Medium particles, at that point, are vibrating with frequency

n H z

3 At that, zero intensity is observed

2 n

times per second

4 None of the above

PHXI15:WAVES

354995 A tuning fork $A$ produces 4 beats $s^{- 1}$ with another tuning fork $B$ of frequency $320 H z .$ On filling one of the prongs of $A, 4$ beats $s^{- 1}$ are again heard when sounded with the same fork $B$ . Then, the frequency of the fork $A$ before filling is

1

328 H z

2

316 H z

3

324 H z

4

320 H z

Explanation:

There are 4 beats between $A$ and $B$ , therefore the possible frequencies of $A$ are 316 or 324 , $i . e .$ $(320 \pm 4) H z$ .
When the prong of $A$ is filled, its frequency becomes greater than the original frequency. If we assume that original frequency of $A$ is 324 , then on filling its frequency will be greater than 324. The beats between $A$ and $B$ will be more than 4.
But it is given that, the beats are again 4, therefore, 324 is not possible.
Therefore, required frequency must be $316 H z .$ This is true because on filing the frequency may increase so as to give 4 beats with $B$ of frequency $320 H z .$

PHXI15:WAVES

354996 Two tuning forks, $A$ and $B$ , produce notes of frequencies $256 H z$ and $262 H z$ . An unknown note sounded with A produces certain beats. When the same note is sounded with $B$ , the beat frequency gets doubled. The unknown frequency is

1

250 H z

2

252 H z

3

254 H z

4

256 H z

Explanation:

$262 - f = 2 (256 - f)$
$f = 250 H z$

PHXI15:WAVES

354997 Two identical flutes produce fundamental notes of frequency $300 H z$ at $27^{\circ} C$ . If the temperature of air in one flute is increased to $31^{\circ} C$ , the number of the beats heard per second will be

1 2

2 1

3 4

4 3

Explanation:

Velocity of sound increases if the temperature increases. So with $v = n λ$ , if $v$ increases $n$ will increse at $27^{\circ} C, v_{1} = n λ, at 31^{\circ} C, v_{2} = (n + x) λ$
Now using $v \propto \sqrt{T} (∵ v = \sqrt{\frac{γ R T}{M}})$
$\Rightarrow \frac{300 + x}{300} = \sqrt{\frac{(273 + 31)}{(273 + 27)}} = \sqrt{\frac{304}{300}} = \sqrt{\frac{300 + 4}{300}}$
$\Rightarrow 1 + \frac{x}{300} = {(1 + \frac{4}{300})}^{1 / 2} = (1 + \frac{1}{2} \times \frac{4}{300}) \Rightarrow x = 2$ .
$[∵ (1 + x)^{n} = 1 + n x]$

PHXI15:WAVES

354998 At a point, beat frequency of $n H z$ is observed. It means:

1 Amplitude of vibrations changes simple harmonically with frequency

n / 2 H z

at that point only

2 Medium particles, at that point, are vibrating with frequency

n H z

3 At that, zero intensity is observed

2 n

times per second

4 None of the above