QBManager

WAVES

173149 If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will

1 increase by a factor of 2

2 decrease by a factor of 2

3 decrease by a factor of 4

4 remains unchanged

Explanation:

C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 .
We know that,
Intensity of sound $(I) = A^{2} n^{2}$
$\frac{I_{1}}{I_{2}} = {(\frac{A_{1}}{A_{2}})}^{2} {(\frac{n_{1}}{n_{2}})}^{2}$
$= {(\frac{A}{2 A})}^{2} {(\frac{4 n}{n})}^{2}$
$I_{2} = 2 A$
$n_{1} = 4 n$
$= \frac{1}{4} \times 16 = 4$
$∴ I_{2} = \frac{I_{1}}{4}$

AIPMT- 1989

WAVES

173150 Velocity of sound waves in air is $330 m / s$ . For a particular sound wave in air, path difference of $40 cm$ is equivalent to phase difference of $1.6$ $π$ . The frequency of this wave is

1

165 Hz

2

150 Hz

3

660 Hz

4

330 Hz

Explanation:

C Given,
Velocity of sound $(v) = 330 m / s$
Path difference $(Δ x) = 40 cm = 0.4 m$
Phase different $(ϕ) = 1.6 π$
We know that,
$\frac{Δ x}{Δ ϕ} = \frac{λ}{2 π}$
Put the value in formula,
$\frac{0.4}{1.6 π} = \frac{λ}{2 π}$
$λ = 0.5 m$
The frequency of wave, $= \frac{V}{λ}$
$= \frac{330 m / s}{0.5}$
$= 660 Hz$

AIPMT-1990

WAVES

173151 In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 s$ . The frequency of the wave is

1

1.47 Hz

2

0.36 Hz

3

0.73 Hz

4

2.94 Hz

Explanation:

A Maximum displacement time (t) $= 0.170 s$
For one vibration,
Time period, $(T) = 4 t = 4 \times 0.17$
$= 0.68 \sec$
Frequency $(f) = \frac{1}{T}$
$= \frac{1}{0.68} = 1.47 Hz$

AIPMT-1998

WAVES

173152 A pipe closed at one end and open at the other end resonates with a sound of frequency $135 Hz$ and also with $165 Hz$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is

1

15 Hz

2

60 Hz

3

7.5 Hz

4

30 Hz

Explanation:

A Given,
The sound frequency $(f_{1}) = 135 Hz$
$f_{2} = 165 Hz$
The different between two frequencies,
$f_{2} - f_{1} = 165 - 135 = 30 Hz$
According to question,
$f_{2} - f_{1} = 2 \times fundamental frequency$
$30 = 2 \times fundamental frequency$
$∴$ Fundamental frequency $= 15 Hz$

AP EAMCET(Medical)-2013

WAVES

173149 If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will

1 increase by a factor of 2

2 decrease by a factor of 2

3 decrease by a factor of 4

4 remains unchanged

Explanation:

C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 .
We know that,
Intensity of sound $(I) = A^{2} n^{2}$
$\frac{I_{1}}{I_{2}} = {(\frac{A_{1}}{A_{2}})}^{2} {(\frac{n_{1}}{n_{2}})}^{2}$
$= {(\frac{A}{2 A})}^{2} {(\frac{4 n}{n})}^{2}$
$I_{2} = 2 A$
$n_{1} = 4 n$
$= \frac{1}{4} \times 16 = 4$
$∴ I_{2} = \frac{I_{1}}{4}$

AIPMT- 1989

WAVES

173150 Velocity of sound waves in air is $330 m / s$ . For a particular sound wave in air, path difference of $40 cm$ is equivalent to phase difference of $1.6$ $π$ . The frequency of this wave is

1

165 Hz

2

150 Hz

3

660 Hz

4

330 Hz

Explanation:

C Given,
Velocity of sound $(v) = 330 m / s$
Path difference $(Δ x) = 40 cm = 0.4 m$
Phase different $(ϕ) = 1.6 π$
We know that,
$\frac{Δ x}{Δ ϕ} = \frac{λ}{2 π}$
Put the value in formula,
$\frac{0.4}{1.6 π} = \frac{λ}{2 π}$
$λ = 0.5 m$
The frequency of wave, $= \frac{V}{λ}$
$= \frac{330 m / s}{0.5}$
$= 660 Hz$

AIPMT-1990

WAVES

173151 In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 s$ . The frequency of the wave is

1

1.47 Hz

2

0.36 Hz

3

0.73 Hz

4

2.94 Hz

Explanation:

A Maximum displacement time (t) $= 0.170 s$
For one vibration,
Time period, $(T) = 4 t = 4 \times 0.17$
$= 0.68 \sec$
Frequency $(f) = \frac{1}{T}$
$= \frac{1}{0.68} = 1.47 Hz$

AIPMT-1998

WAVES

173152 A pipe closed at one end and open at the other end resonates with a sound of frequency $135 Hz$ and also with $165 Hz$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is

1

15 Hz

2

60 Hz

3

7.5 Hz

4

30 Hz

Explanation:

A Given,
The sound frequency $(f_{1}) = 135 Hz$
$f_{2} = 165 Hz$
The different between two frequencies,
$f_{2} - f_{1} = 165 - 135 = 30 Hz$
According to question,
$f_{2} - f_{1} = 2 \times fundamental frequency$
$30 = 2 \times fundamental frequency$
$∴$ Fundamental frequency $= 15 Hz$

AP EAMCET(Medical)-2013

WAVES

173149 If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will

1 increase by a factor of 2

2 decrease by a factor of 2

3 decrease by a factor of 4

4 remains unchanged

Explanation:

C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 .
We know that,
Intensity of sound $(I) = A^{2} n^{2}$
$\frac{I_{1}}{I_{2}} = {(\frac{A_{1}}{A_{2}})}^{2} {(\frac{n_{1}}{n_{2}})}^{2}$
$= {(\frac{A}{2 A})}^{2} {(\frac{4 n}{n})}^{2}$
$I_{2} = 2 A$
$n_{1} = 4 n$
$= \frac{1}{4} \times 16 = 4$
$∴ I_{2} = \frac{I_{1}}{4}$

AIPMT- 1989

WAVES

173150 Velocity of sound waves in air is $330 m / s$ . For a particular sound wave in air, path difference of $40 cm$ is equivalent to phase difference of $1.6$ $π$ . The frequency of this wave is

1

165 Hz

2

150 Hz

3

660 Hz

4

330 Hz

Explanation:

C Given,
Velocity of sound $(v) = 330 m / s$
Path difference $(Δ x) = 40 cm = 0.4 m$
Phase different $(ϕ) = 1.6 π$
We know that,
$\frac{Δ x}{Δ ϕ} = \frac{λ}{2 π}$
Put the value in formula,
$\frac{0.4}{1.6 π} = \frac{λ}{2 π}$
$λ = 0.5 m$
The frequency of wave, $= \frac{V}{λ}$
$= \frac{330 m / s}{0.5}$
$= 660 Hz$

AIPMT-1990

WAVES

173151 In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 s$ . The frequency of the wave is

1

1.47 Hz

2

0.36 Hz

3

0.73 Hz

4

2.94 Hz

Explanation:

A Maximum displacement time (t) $= 0.170 s$
For one vibration,
Time period, $(T) = 4 t = 4 \times 0.17$
$= 0.68 \sec$
Frequency $(f) = \frac{1}{T}$
$= \frac{1}{0.68} = 1.47 Hz$

AIPMT-1998

WAVES

173152 A pipe closed at one end and open at the other end resonates with a sound of frequency $135 Hz$ and also with $165 Hz$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is

1

15 Hz

2

60 Hz

3

7.5 Hz

4

30 Hz

Explanation:

A Given,
The sound frequency $(f_{1}) = 135 Hz$
$f_{2} = 165 Hz$
The different between two frequencies,
$f_{2} - f_{1} = 165 - 135 = 30 Hz$
According to question,
$f_{2} - f_{1} = 2 \times fundamental frequency$
$30 = 2 \times fundamental frequency$
$∴$ Fundamental frequency $= 15 Hz$

AP EAMCET(Medical)-2013

WAVES

173149 If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will

1 increase by a factor of 2

2 decrease by a factor of 2

3 decrease by a factor of 4

4 remains unchanged

Explanation:

C If the amplitude of sound is doubled and the frequency reduced to one-fourth, the intensity of sound at the same point will decrease by a factor of 4 .
We know that,
Intensity of sound $(I) = A^{2} n^{2}$
$\frac{I_{1}}{I_{2}} = {(\frac{A_{1}}{A_{2}})}^{2} {(\frac{n_{1}}{n_{2}})}^{2}$
$= {(\frac{A}{2 A})}^{2} {(\frac{4 n}{n})}^{2}$
$I_{2} = 2 A$
$n_{1} = 4 n$
$= \frac{1}{4} \times 16 = 4$
$∴ I_{2} = \frac{I_{1}}{4}$

AIPMT- 1989

WAVES

173150 Velocity of sound waves in air is $330 m / s$ . For a particular sound wave in air, path difference of $40 cm$ is equivalent to phase difference of $1.6$ $π$ . The frequency of this wave is

1

165 Hz

2

150 Hz

3

660 Hz

4

330 Hz

Explanation:

C Given,
Velocity of sound $(v) = 330 m / s$
Path difference $(Δ x) = 40 cm = 0.4 m$
Phase different $(ϕ) = 1.6 π$
We know that,
$\frac{Δ x}{Δ ϕ} = \frac{λ}{2 π}$
Put the value in formula,
$\frac{0.4}{1.6 π} = \frac{λ}{2 π}$
$λ = 0.5 m$
The frequency of wave, $= \frac{V}{λ}$
$= \frac{330 m / s}{0.5}$
$= 660 Hz$

AIPMT-1990

WAVES

173151 In a sinusoidal wave, the time required for a particular point, to move from maximum displacement to zero displacement is $0.170 s$ . The frequency of the wave is

1

1.47 Hz

2

0.36 Hz

3

0.73 Hz

4

2.94 Hz

Explanation:

A Maximum displacement time (t) $= 0.170 s$
For one vibration,
Time period, $(T) = 4 t = 4 \times 0.17$
$= 0.68 \sec$
Frequency $(f) = \frac{1}{T}$
$= \frac{1}{0.68} = 1.47 Hz$

AIPMT-1998

WAVES

173152 A pipe closed at one end and open at the other end resonates with a sound of frequency $135 Hz$ and also with $165 Hz$ but not at any other frequency intermediate between these two. Then the frequency of the fundamental note of the pipe is

1

15 Hz

2

60 Hz

3

7.5 Hz

4

30 Hz

Explanation:

A Given,
The sound frequency $(f_{1}) = 135 Hz$
$f_{2} = 165 Hz$
The different between two frequencies,
$f_{2} - f_{1} = 165 - 135 = 30 Hz$
According to question,
$f_{2} - f_{1} = 2 \times fundamental frequency$
$30 = 2 \times fundamental frequency$
$∴$ Fundamental frequency $= 15 Hz$

AP EAMCET(Medical)-2013

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