QBManager

WAVES

172772 The beat frequency observed when two sound waves $y_{1} = 0.5 \sin (410 t)$ and $y_{2} = 0.5 \sin$ $(454 t)$ travel in the same direction is

1 5

2 3

3 7

4 2

5 4

Explanation:

C Given that,
Equations of waves are-
$y_{1} = 0.5 \sin (410 t)$
$y_{2} = 0.5 \sin (454 t)$
Comparing above equations with standard equation
$y = a \sin (ω t - kx) we get$
$ω_{1} = 410 & ω_{2} = 454$
Now, frequency of
$f_{1} = \frac{410}{2 π} and f_{2} = \frac{454}{2 π}$
$f_{1} = \frac{205 \times 7}{22} f_{2} = \frac{227 \times 7}{22}$
$f_{1} = 65.227 Hz f_{2} = 72.227 Hz$
Hence, the beat frequency $(f) = f_{1} - f_{2}$
$f = 72.227 - 65.227 ⊔ 7 beats$

Kerala CEE 2012

WAVES

172773 Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1} = 4 \sin 600 π t$ and $Y_{2} = 5 \sin 608 π t$
An observer located near these two sources of sound will hear

1 4 beat/s with intensity ratio

25 : 16

between waxing and waning

2 8 beat/s with intensity ratio

25 : 16

between waxing and waning

3 8 beat/s with intensity ratio

81 : 1

between waxing and waning

4 4 beat/s with intensity ratio

81 : 1

between waxing and waning

Explanation:

D Given progressive wave equation
$Y_{1} = 4 \sin 600 π t$
$Y_{2} = 5 \sin 608 π t$
$standard equation$
$Y = A \sin ω t$
$from equation (i)$
$ω t == 600 π t$
$2 π f_{1} t = 600 π t$
$f_{1} = 300 Hz$
$from equation (ii)$
$2 π f_{2} t = 608 π t$
$f_{2} = 304 Hz$
Number of beats $= f_{2} - f_{1} = 304 - 300 = 4$ beats $/ \sec$
$\frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{4 + 5}{4 - 5})}^{2}$
$I_{max} : I_{min} = 81 : 1$

AIPMT-2012

WAVES

172777 There are two waves having wavelengths 100 $cm$ and $101 cm$ and same velocity $303 m / s$ . The beat frequency is

1

4 Hz

2

1 Hz

3

3 Hz

4

2 Hz

Explanation:

C Given,
$λ_{1} = 100 cm$
$λ_{2} = 101 cm$
Velocity, $v = 303 m / s$
Beat frequency, $f_{b} = f_{1} - f_{2}$
$f_{b} = (\frac{v}{λ_{1}} - \frac{v}{λ_{2}})$
$= \frac{303 \times 10^{2}}{100} - \frac{303 \times 10^{2}}{101}$
$= 3.03 \times 10^{2} - 3 \times 10^{2}$
$= 3 Hz$

UPSEE - 2016

WAVES

172842 The intensity of sound gets reduced by $10 %$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is

1

30 %

2

27.1 %

3

20 %

4

36 %

Explanation:

B After passing through $1^{st}$ slab,
Intensity $= 90 %$ of $I = 0.9 I$
After second slab, intensity $= 90 %$ of $(0.9 I)$
$= 81 % I = 0.81 I$
After third slab, intensity $= 90 %$ of $(0.81 I)$
$= 72.9 % of I$
$∴$ Reduction in intensity $= 27.1 %$

J and K CET- 2004

WAVES

172779 A sound absorber attenuates the sound level by $20 dB$ . The intensity decreases by a factor of

1 1000

2 10000

3 10

4 100

Explanation:

D The intensity level of the sound initially will be ,
$β_{1} = 10 \log (\frac{I}{I_{0}})$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be -
$β_{2} = 10 \log (\frac{I^{'}}{I_{0}})$
As it is given that the sound is decreased by $20 dB$ , which means that the difference between initial sound intensity level and final sound intensity level is $20 dB$ , Therefore
$β_{1} - β_{2} = 10 \log (\frac{I}{I_{\circ}}) - 10 \log (\frac{I^{'}}{I_{\circ}})$
$20 = 10 \log (\frac{I}{I^{'}})$
$2 = \log (\frac{I}{I^{'}})$
$- 2 = \log (\frac{I^{'}}{I})$
$10^{- 2} = \frac{I^{'}}{I}$
$I^{'} = \frac{I}{100}$
Therefore, the intensity of sound decreased by a factor of 100 .

UPSEE - 2007

WAVES

172772 The beat frequency observed when two sound waves $y_{1} = 0.5 \sin (410 t)$ and $y_{2} = 0.5 \sin$ $(454 t)$ travel in the same direction is

1 5

2 3

3 7

4 2

5 4

Explanation:

C Given that,
Equations of waves are-
$y_{1} = 0.5 \sin (410 t)$
$y_{2} = 0.5 \sin (454 t)$
Comparing above equations with standard equation
$y = a \sin (ω t - kx) we get$
$ω_{1} = 410 & ω_{2} = 454$
Now, frequency of
$f_{1} = \frac{410}{2 π} and f_{2} = \frac{454}{2 π}$
$f_{1} = \frac{205 \times 7}{22} f_{2} = \frac{227 \times 7}{22}$
$f_{1} = 65.227 Hz f_{2} = 72.227 Hz$
Hence, the beat frequency $(f) = f_{1} - f_{2}$
$f = 72.227 - 65.227 ⊔ 7 beats$

Kerala CEE 2012

WAVES

172773 Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1} = 4 \sin 600 π t$ and $Y_{2} = 5 \sin 608 π t$
An observer located near these two sources of sound will hear

1 4 beat/s with intensity ratio

25 : 16

between waxing and waning

2 8 beat/s with intensity ratio

25 : 16

between waxing and waning

3 8 beat/s with intensity ratio

81 : 1

between waxing and waning

4 4 beat/s with intensity ratio

81 : 1

between waxing and waning

Explanation:

D Given progressive wave equation
$Y_{1} = 4 \sin 600 π t$
$Y_{2} = 5 \sin 608 π t$
$standard equation$
$Y = A \sin ω t$
$from equation (i)$
$ω t == 600 π t$
$2 π f_{1} t = 600 π t$
$f_{1} = 300 Hz$
$from equation (ii)$
$2 π f_{2} t = 608 π t$
$f_{2} = 304 Hz$
Number of beats $= f_{2} - f_{1} = 304 - 300 = 4$ beats $/ \sec$
$\frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{4 + 5}{4 - 5})}^{2}$
$I_{max} : I_{min} = 81 : 1$

AIPMT-2012

WAVES

172777 There are two waves having wavelengths 100 $cm$ and $101 cm$ and same velocity $303 m / s$ . The beat frequency is

1

4 Hz

2

1 Hz

3

3 Hz

4

2 Hz

Explanation:

C Given,
$λ_{1} = 100 cm$
$λ_{2} = 101 cm$
Velocity, $v = 303 m / s$
Beat frequency, $f_{b} = f_{1} - f_{2}$
$f_{b} = (\frac{v}{λ_{1}} - \frac{v}{λ_{2}})$
$= \frac{303 \times 10^{2}}{100} - \frac{303 \times 10^{2}}{101}$
$= 3.03 \times 10^{2} - 3 \times 10^{2}$
$= 3 Hz$

UPSEE - 2016

WAVES

172842 The intensity of sound gets reduced by $10 %$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is

1

30 %

2

27.1 %

3

20 %

4

36 %

Explanation:

B After passing through $1^{st}$ slab,
Intensity $= 90 %$ of $I = 0.9 I$
After second slab, intensity $= 90 %$ of $(0.9 I)$
$= 81 % I = 0.81 I$
After third slab, intensity $= 90 %$ of $(0.81 I)$
$= 72.9 % of I$
$∴$ Reduction in intensity $= 27.1 %$

J and K CET- 2004

WAVES

172779 A sound absorber attenuates the sound level by $20 dB$ . The intensity decreases by a factor of

1 1000

2 10000

3 10

4 100

Explanation:

D The intensity level of the sound initially will be ,
$β_{1} = 10 \log (\frac{I}{I_{0}})$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be -
$β_{2} = 10 \log (\frac{I^{'}}{I_{0}})$
As it is given that the sound is decreased by $20 dB$ , which means that the difference between initial sound intensity level and final sound intensity level is $20 dB$ , Therefore
$β_{1} - β_{2} = 10 \log (\frac{I}{I_{\circ}}) - 10 \log (\frac{I^{'}}{I_{\circ}})$
$20 = 10 \log (\frac{I}{I^{'}})$
$2 = \log (\frac{I}{I^{'}})$
$- 2 = \log (\frac{I^{'}}{I})$
$10^{- 2} = \frac{I^{'}}{I}$
$I^{'} = \frac{I}{100}$
Therefore, the intensity of sound decreased by a factor of 100 .

UPSEE - 2007

WAVES

172772 The beat frequency observed when two sound waves $y_{1} = 0.5 \sin (410 t)$ and $y_{2} = 0.5 \sin$ $(454 t)$ travel in the same direction is

1 5

2 3

3 7

4 2

5 4

Explanation:

C Given that,
Equations of waves are-
$y_{1} = 0.5 \sin (410 t)$
$y_{2} = 0.5 \sin (454 t)$
Comparing above equations with standard equation
$y = a \sin (ω t - kx) we get$
$ω_{1} = 410 & ω_{2} = 454$
Now, frequency of
$f_{1} = \frac{410}{2 π} and f_{2} = \frac{454}{2 π}$
$f_{1} = \frac{205 \times 7}{22} f_{2} = \frac{227 \times 7}{22}$
$f_{1} = 65.227 Hz f_{2} = 72.227 Hz$
Hence, the beat frequency $(f) = f_{1} - f_{2}$
$f = 72.227 - 65.227 ⊔ 7 beats$

Kerala CEE 2012

WAVES

172773 Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1} = 4 \sin 600 π t$ and $Y_{2} = 5 \sin 608 π t$
An observer located near these two sources of sound will hear

1 4 beat/s with intensity ratio

25 : 16

between waxing and waning

2 8 beat/s with intensity ratio

25 : 16

between waxing and waning

3 8 beat/s with intensity ratio

81 : 1

between waxing and waning

4 4 beat/s with intensity ratio

81 : 1

between waxing and waning

Explanation:

D Given progressive wave equation
$Y_{1} = 4 \sin 600 π t$
$Y_{2} = 5 \sin 608 π t$
$standard equation$
$Y = A \sin ω t$
$from equation (i)$
$ω t == 600 π t$
$2 π f_{1} t = 600 π t$
$f_{1} = 300 Hz$
$from equation (ii)$
$2 π f_{2} t = 608 π t$
$f_{2} = 304 Hz$
Number of beats $= f_{2} - f_{1} = 304 - 300 = 4$ beats $/ \sec$
$\frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{4 + 5}{4 - 5})}^{2}$
$I_{max} : I_{min} = 81 : 1$

AIPMT-2012

WAVES

172777 There are two waves having wavelengths 100 $cm$ and $101 cm$ and same velocity $303 m / s$ . The beat frequency is

1

4 Hz

2

1 Hz

3

3 Hz

4

2 Hz

Explanation:

C Given,
$λ_{1} = 100 cm$
$λ_{2} = 101 cm$
Velocity, $v = 303 m / s$
Beat frequency, $f_{b} = f_{1} - f_{2}$
$f_{b} = (\frac{v}{λ_{1}} - \frac{v}{λ_{2}})$
$= \frac{303 \times 10^{2}}{100} - \frac{303 \times 10^{2}}{101}$
$= 3.03 \times 10^{2} - 3 \times 10^{2}$
$= 3 Hz$

UPSEE - 2016

WAVES

172842 The intensity of sound gets reduced by $10 %$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is

1

30 %

2

27.1 %

3

20 %

4

36 %

Explanation:

B After passing through $1^{st}$ slab,
Intensity $= 90 %$ of $I = 0.9 I$
After second slab, intensity $= 90 %$ of $(0.9 I)$
$= 81 % I = 0.81 I$
After third slab, intensity $= 90 %$ of $(0.81 I)$
$= 72.9 % of I$
$∴$ Reduction in intensity $= 27.1 %$

J and K CET- 2004

WAVES

172779 A sound absorber attenuates the sound level by $20 dB$ . The intensity decreases by a factor of

1 1000

2 10000

3 10

4 100

Explanation:

D The intensity level of the sound initially will be ,
$β_{1} = 10 \log (\frac{I}{I_{0}})$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be -
$β_{2} = 10 \log (\frac{I^{'}}{I_{0}})$
As it is given that the sound is decreased by $20 dB$ , which means that the difference between initial sound intensity level and final sound intensity level is $20 dB$ , Therefore
$β_{1} - β_{2} = 10 \log (\frac{I}{I_{\circ}}) - 10 \log (\frac{I^{'}}{I_{\circ}})$
$20 = 10 \log (\frac{I}{I^{'}})$
$2 = \log (\frac{I}{I^{'}})$
$- 2 = \log (\frac{I^{'}}{I})$
$10^{- 2} = \frac{I^{'}}{I}$
$I^{'} = \frac{I}{100}$
Therefore, the intensity of sound decreased by a factor of 100 .

UPSEE - 2007

WAVES

172772 The beat frequency observed when two sound waves $y_{1} = 0.5 \sin (410 t)$ and $y_{2} = 0.5 \sin$ $(454 t)$ travel in the same direction is

1 5

2 3

3 7

4 2

5 4

Explanation:

C Given that,
Equations of waves are-
$y_{1} = 0.5 \sin (410 t)$
$y_{2} = 0.5 \sin (454 t)$
Comparing above equations with standard equation
$y = a \sin (ω t - kx) we get$
$ω_{1} = 410 & ω_{2} = 454$
Now, frequency of
$f_{1} = \frac{410}{2 π} and f_{2} = \frac{454}{2 π}$
$f_{1} = \frac{205 \times 7}{22} f_{2} = \frac{227 \times 7}{22}$
$f_{1} = 65.227 Hz f_{2} = 72.227 Hz$
Hence, the beat frequency $(f) = f_{1} - f_{2}$
$f = 72.227 - 65.227 ⊔ 7 beats$

Kerala CEE 2012

WAVES

172773 Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1} = 4 \sin 600 π t$ and $Y_{2} = 5 \sin 608 π t$
An observer located near these two sources of sound will hear

1 4 beat/s with intensity ratio

25 : 16

between waxing and waning

2 8 beat/s with intensity ratio

25 : 16

between waxing and waning

3 8 beat/s with intensity ratio

81 : 1

between waxing and waning

4 4 beat/s with intensity ratio

81 : 1

between waxing and waning

Explanation:

D Given progressive wave equation
$Y_{1} = 4 \sin 600 π t$
$Y_{2} = 5 \sin 608 π t$
$standard equation$
$Y = A \sin ω t$
$from equation (i)$
$ω t == 600 π t$
$2 π f_{1} t = 600 π t$
$f_{1} = 300 Hz$
$from equation (ii)$
$2 π f_{2} t = 608 π t$
$f_{2} = 304 Hz$
Number of beats $= f_{2} - f_{1} = 304 - 300 = 4$ beats $/ \sec$
$\frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{4 + 5}{4 - 5})}^{2}$
$I_{max} : I_{min} = 81 : 1$

AIPMT-2012

WAVES

172777 There are two waves having wavelengths 100 $cm$ and $101 cm$ and same velocity $303 m / s$ . The beat frequency is

1

4 Hz

2

1 Hz

3

3 Hz

4

2 Hz

Explanation:

C Given,
$λ_{1} = 100 cm$
$λ_{2} = 101 cm$
Velocity, $v = 303 m / s$
Beat frequency, $f_{b} = f_{1} - f_{2}$
$f_{b} = (\frac{v}{λ_{1}} - \frac{v}{λ_{2}})$
$= \frac{303 \times 10^{2}}{100} - \frac{303 \times 10^{2}}{101}$
$= 3.03 \times 10^{2} - 3 \times 10^{2}$
$= 3 Hz$

UPSEE - 2016

WAVES

172842 The intensity of sound gets reduced by $10 %$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is

1

30 %

2

27.1 %

3

20 %

4

36 %

Explanation:

B After passing through $1^{st}$ slab,
Intensity $= 90 %$ of $I = 0.9 I$
After second slab, intensity $= 90 %$ of $(0.9 I)$
$= 81 % I = 0.81 I$
After third slab, intensity $= 90 %$ of $(0.81 I)$
$= 72.9 % of I$
$∴$ Reduction in intensity $= 27.1 %$

J and K CET- 2004

WAVES

172779 A sound absorber attenuates the sound level by $20 dB$ . The intensity decreases by a factor of

1 1000

2 10000

3 10

4 100

Explanation:

D The intensity level of the sound initially will be ,
$β_{1} = 10 \log (\frac{I}{I_{0}})$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be -
$β_{2} = 10 \log (\frac{I^{'}}{I_{0}})$
As it is given that the sound is decreased by $20 dB$ , which means that the difference between initial sound intensity level and final sound intensity level is $20 dB$ , Therefore
$β_{1} - β_{2} = 10 \log (\frac{I}{I_{\circ}}) - 10 \log (\frac{I^{'}}{I_{\circ}})$
$20 = 10 \log (\frac{I}{I^{'}})$
$2 = \log (\frac{I}{I^{'}})$
$- 2 = \log (\frac{I^{'}}{I})$
$10^{- 2} = \frac{I^{'}}{I}$
$I^{'} = \frac{I}{100}$
Therefore, the intensity of sound decreased by a factor of 100 .

UPSEE - 2007

WAVES

172772 The beat frequency observed when two sound waves $y_{1} = 0.5 \sin (410 t)$ and $y_{2} = 0.5 \sin$ $(454 t)$ travel in the same direction is

1 5

2 3

3 7

4 2

5 4

Explanation:

C Given that,
Equations of waves are-
$y_{1} = 0.5 \sin (410 t)$
$y_{2} = 0.5 \sin (454 t)$
Comparing above equations with standard equation
$y = a \sin (ω t - kx) we get$
$ω_{1} = 410 & ω_{2} = 454$
Now, frequency of
$f_{1} = \frac{410}{2 π} and f_{2} = \frac{454}{2 π}$
$f_{1} = \frac{205 \times 7}{22} f_{2} = \frac{227 \times 7}{22}$
$f_{1} = 65.227 Hz f_{2} = 72.227 Hz$
Hence, the beat frequency $(f) = f_{1} - f_{2}$
$f = 72.227 - 65.227 ⊔ 7 beats$

Kerala CEE 2012

WAVES

172773 Two sources of sound placed closed to each other, are emitting progressive waves given by $Y_{1} = 4 \sin 600 π t$ and $Y_{2} = 5 \sin 608 π t$
An observer located near these two sources of sound will hear

1 4 beat/s with intensity ratio

25 : 16

between waxing and waning

2 8 beat/s with intensity ratio

25 : 16

between waxing and waning

3 8 beat/s with intensity ratio

81 : 1

between waxing and waning

4 4 beat/s with intensity ratio

81 : 1

between waxing and waning

Explanation:

D Given progressive wave equation
$Y_{1} = 4 \sin 600 π t$
$Y_{2} = 5 \sin 608 π t$
$standard equation$
$Y = A \sin ω t$
$from equation (i)$
$ω t == 600 π t$
$2 π f_{1} t = 600 π t$
$f_{1} = 300 Hz$
$from equation (ii)$
$2 π f_{2} t = 608 π t$
$f_{2} = 304 Hz$
Number of beats $= f_{2} - f_{1} = 304 - 300 = 4$ beats $/ \sec$
$\frac{I_{max}}{I_{min}} = {(\frac{A_{1} + A_{2}}{A_{1} - A_{2}})}^{2}$
$\frac{I_{max}}{I_{min}} = {(\frac{4 + 5}{4 - 5})}^{2}$
$I_{max} : I_{min} = 81 : 1$

AIPMT-2012

WAVES

172777 There are two waves having wavelengths 100 $cm$ and $101 cm$ and same velocity $303 m / s$ . The beat frequency is

1

4 Hz

2

1 Hz

3

3 Hz

4

2 Hz

Explanation:

C Given,
$λ_{1} = 100 cm$
$λ_{2} = 101 cm$
Velocity, $v = 303 m / s$
Beat frequency, $f_{b} = f_{1} - f_{2}$
$f_{b} = (\frac{v}{λ_{1}} - \frac{v}{λ_{2}})$
$= \frac{303 \times 10^{2}}{100} - \frac{303 \times 10^{2}}{101}$
$= 3.03 \times 10^{2} - 3 \times 10^{2}$
$= 3 Hz$

UPSEE - 2016

WAVES

172842 The intensity of sound gets reduced by $10 %$ on passing through a slab. The reduction in intensity on passage through three consecutive slabs is

1

30 %

2

27.1 %

3

20 %

4

36 %

Explanation:

B After passing through $1^{st}$ slab,
Intensity $= 90 %$ of $I = 0.9 I$
After second slab, intensity $= 90 %$ of $(0.9 I)$
$= 81 % I = 0.81 I$
After third slab, intensity $= 90 %$ of $(0.81 I)$
$= 72.9 % of I$
$∴$ Reduction in intensity $= 27.1 %$

J and K CET- 2004

WAVES

172779 A sound absorber attenuates the sound level by $20 dB$ . The intensity decreases by a factor of

1 1000

2 10000

3 10

4 100

Explanation:

D The intensity level of the sound initially will be ,
$β_{1} = 10 \log (\frac{I}{I_{0}})$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be -
$β_{2} = 10 \log (\frac{I^{'}}{I_{0}})$
As it is given that the sound is decreased by $20 dB$ , which means that the difference between initial sound intensity level and final sound intensity level is $20 dB$ , Therefore
$β_{1} - β_{2} = 10 \log (\frac{I}{I_{\circ}}) - 10 \log (\frac{I^{'}}{I_{\circ}})$
$20 = 10 \log (\frac{I}{I^{'}})$
$2 = \log (\frac{I}{I^{'}})$
$- 2 = \log (\frac{I^{'}}{I})$
$10^{- 2} = \frac{I^{'}}{I}$
$I^{'} = \frac{I}{100}$
Therefore, the intensity of sound decreased by a factor of 100 .

UPSEE - 2007

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