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PHXI15:WAVES

354812 The ratio of intensities between two coherent sound sources is $4 : 1$ . The difference of loudness in decibels $(dB)$ between maximum and minimum intensities when they interfere in space is:

1

10 \log 2

2

20 \log 3

3

10 \log 3

4

20 \log 2

Explanation:

Given that $\frac{I_{1}}{I_{2}} = \frac{4}{1}$
$∴ \frac{I_{max}}{I_{max}} = {[\frac{\sqrt{I_{1} / I_{2}} + 1}{\sqrt{I_{1} / I_{2}} - 1}]}^{2} = {[\frac{2 + 1}{2 - 1}]}^{2} = 9$
The difference of loudness in decibles is
$\begin{array}{r} β_{1} - β_{2} = \log \frac{I_{max}}{I_{0}} - 10 \log \frac{I_{min}}{I_{0}} \\ = 10 \log \frac{I_{max}}{I_{min}} = 10 \log (9) = 20 \log (3) \end{array}$

PHXI15:WAVES

354813 If $T$ is the reverberation time of an auditorium of volume $V$ then

1

T \propto \frac{1}{V^{2}}

2

T \propto \frac{1}{V}

3

T \propto V

4

T \propto V^{2}

Explanation:

Reverberation time $T \propto V$ .

PHXI15:WAVES

354814 The intensity of sound increases at night due to:

1 Increases in density of air

2 Decreases in density of air

3 High temperature

4 None of these

PHXI15:WAVES

354815 The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be

1 16

2 8

3 14

4 7

Explanation:

$β = 10 \log \frac{I}{I_{0}}$
$\begin{aligned} β_{1} = 4 = 10 \log \frac{I}{I_{0}} \\ β_{2} = 10 \log (\frac{2 I}{I_{0}}) = 4 + 10 \log (2) = 7 \end{aligned}$

PHXI15:WAVES

354812 The ratio of intensities between two coherent sound sources is $4 : 1$ . The difference of loudness in decibels $(dB)$ between maximum and minimum intensities when they interfere in space is:

1

10 \log 2

2

20 \log 3

3

10 \log 3

4

20 \log 2

Explanation:

Given that $\frac{I_{1}}{I_{2}} = \frac{4}{1}$
$∴ \frac{I_{max}}{I_{max}} = {[\frac{\sqrt{I_{1} / I_{2}} + 1}{\sqrt{I_{1} / I_{2}} - 1}]}^{2} = {[\frac{2 + 1}{2 - 1}]}^{2} = 9$
The difference of loudness in decibles is
$\begin{array}{r} β_{1} - β_{2} = \log \frac{I_{max}}{I_{0}} - 10 \log \frac{I_{min}}{I_{0}} \\ = 10 \log \frac{I_{max}}{I_{min}} = 10 \log (9) = 20 \log (3) \end{array}$

PHXI15:WAVES

354813 If $T$ is the reverberation time of an auditorium of volume $V$ then

1

T \propto \frac{1}{V^{2}}

2

T \propto \frac{1}{V}

3

T \propto V

4

T \propto V^{2}

Explanation:

Reverberation time $T \propto V$ .

PHXI15:WAVES

354814 The intensity of sound increases at night due to:

1 Increases in density of air

2 Decreases in density of air

3 High temperature

4 None of these

PHXI15:WAVES

354815 The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be

1 16

2 8

3 14

4 7

Explanation:

$β = 10 \log \frac{I}{I_{0}}$
$\begin{aligned} β_{1} = 4 = 10 \log \frac{I}{I_{0}} \\ β_{2} = 10 \log (\frac{2 I}{I_{0}}) = 4 + 10 \log (2) = 7 \end{aligned}$

PHXI15:WAVES

354812 The ratio of intensities between two coherent sound sources is $4 : 1$ . The difference of loudness in decibels $(dB)$ between maximum and minimum intensities when they interfere in space is:

1

10 \log 2

2

20 \log 3

3

10 \log 3

4

20 \log 2

Explanation:

Given that $\frac{I_{1}}{I_{2}} = \frac{4}{1}$
$∴ \frac{I_{max}}{I_{max}} = {[\frac{\sqrt{I_{1} / I_{2}} + 1}{\sqrt{I_{1} / I_{2}} - 1}]}^{2} = {[\frac{2 + 1}{2 - 1}]}^{2} = 9$
The difference of loudness in decibles is
$\begin{array}{r} β_{1} - β_{2} = \log \frac{I_{max}}{I_{0}} - 10 \log \frac{I_{min}}{I_{0}} \\ = 10 \log \frac{I_{max}}{I_{min}} = 10 \log (9) = 20 \log (3) \end{array}$

PHXI15:WAVES

354813 If $T$ is the reverberation time of an auditorium of volume $V$ then

1

T \propto \frac{1}{V^{2}}

2

T \propto \frac{1}{V}

3

T \propto V

4

T \propto V^{2}

Explanation:

Reverberation time $T \propto V$ .

PHXI15:WAVES

354814 The intensity of sound increases at night due to:

1 Increases in density of air

2 Decreases in density of air

3 High temperature

4 None of these

PHXI15:WAVES

354815 The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be

1 16

2 8

3 14

4 7

Explanation:

$β = 10 \log \frac{I}{I_{0}}$
$\begin{aligned} β_{1} = 4 = 10 \log \frac{I}{I_{0}} \\ β_{2} = 10 \log (\frac{2 I}{I_{0}}) = 4 + 10 \log (2) = 7 \end{aligned}$

PHXI15:WAVES

354812 The ratio of intensities between two coherent sound sources is $4 : 1$ . The difference of loudness in decibels $(dB)$ between maximum and minimum intensities when they interfere in space is:

1

10 \log 2

2

20 \log 3

3

10 \log 3

4

20 \log 2

Explanation:

Given that $\frac{I_{1}}{I_{2}} = \frac{4}{1}$
$∴ \frac{I_{max}}{I_{max}} = {[\frac{\sqrt{I_{1} / I_{2}} + 1}{\sqrt{I_{1} / I_{2}} - 1}]}^{2} = {[\frac{2 + 1}{2 - 1}]}^{2} = 9$
The difference of loudness in decibles is
$\begin{array}{r} β_{1} - β_{2} = \log \frac{I_{max}}{I_{0}} - 10 \log \frac{I_{min}}{I_{0}} \\ = 10 \log \frac{I_{max}}{I_{min}} = 10 \log (9) = 20 \log (3) \end{array}$

PHXI15:WAVES

354813 If $T$ is the reverberation time of an auditorium of volume $V$ then

1

T \propto \frac{1}{V^{2}}

2

T \propto \frac{1}{V}

3

T \propto V

4

T \propto V^{2}

Explanation:

Reverberation time $T \propto V$ .

PHXI15:WAVES

354814 The intensity of sound increases at night due to:

1 Increases in density of air

2 Decreases in density of air

3 High temperature

4 None of these

PHXI15:WAVES

354815 The intensity level of sound wave is said to be 4 decibel. If the intensity of wave is doubled, then the intensity level of sound as expressed in decibel would be

1 16

2 8

3 14

4 7

Explanation:

$β = 10 \log \frac{I}{I_{0}}$
$\begin{aligned} β_{1} = 4 = 10 \log \frac{I}{I_{0}} \\ β_{2} = 10 \log (\frac{2 I}{I_{0}}) = 4 + 10 \log (2) = 7 \end{aligned}$

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