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PHXI15:WAVES

354702 A bat moving at $10 m s^{- 1}$ towards a wall sends a sound signal of $8000 H z$ towards it. On refelection it hears a sound of frequency $f$ . The value of $f$ in $H z$ is close to (speed of sound $= 320 m s^{- 1}$ )

1 8516

2 8258

3 8424

4 8000

Explanation:

Reflected frequency of sound reaching bat
$= [\frac{v - (- v_{0})}{v - v_{s}}] f = [\frac{v + v_{0}}{v - v_{s}}] f = \frac{v + 10}{v - 10} f$
$= (\frac{320 + 10}{320 - 10}) \times 8000 = 8516 H z$

JEE - 2014

PHXI15:WAVES

354703 A boy is walking away from a wall at a speed of $1.0 m / s$ as shown in the figure. The boy blows a whistle steadily. An observer towards whom the boy is moving hears 4 beats. If the speed of sound is $340 m / s$ , the frequency of whistle is:
supporting img

1

480 H z

2

680 H z

3

840 H z

4

100 H z

Explanation:

Direct frequency received by the observer is
$f_{1} = f (\frac{340}{340 - 1}) (1)$
Frequency received by the wall is
$f_{2} = f (\frac{340}{340 + 1}) (2)$
Frequency received by the observer after reflection from the wall is
$f_{3} = f_{2}$
Given that
$f_{1} \sim f_{3} = 4$
$f [\frac{340}{339} - \frac{340}{341}] = 4$
$\Rightarrow f = 680 H z$

PHXI15:WAVES

354704 A person with vibrating tuning fork of frequency $338 H z$ is moving towards a vertical wall with a speed of $2 m s^{- 1}$ . Velocity of sound in air is 340 $m s^{- 1}$ . The number of beats heard by that person per second is

1 2

2 4

3 6

4 8

Explanation:

Frequency of tuning fork (source), $f = 338 H z$
Velocity of sound, $v = 340 m s^{- 1}$
When the person is moving towards the wall, the frequency of sound reaching the wall is
$f^{'} = \frac{f v}{v - v_{S}} (1)$
Where $v_{s} =$ speed of the person $= 2 m s^{- 1}$ .
On reflection the wall acts as source and the person is the listener moving towards the source. So the frequency heard by the person after reflection from the wall is
$f^{''} = [\frac{v + v_{L}}{v}] f^{'} (2)$
Substituting the value of $f^{'}$ from eq. (1) in (2), we get
$f^{''} = [\frac{v + v_{L}}{v - v_{S}}] f = (\frac{v + v_{S}}{v - v_{S}}) f [A s v_{L} = v_{S}]$
$= [\frac{340 + 2}{340 - 2}] \times 338 = 342 H z$
$∴$ Number of beats heard per second $f^{''} - f = 342 - 338 = 4$

KCET - 2013

PHXI15:WAVES

354705 supporting img
A source s emitting sound of $400 H z$ is fixed on block $A$ which is attached to free end of a spring $S_{A}$ . The detector $D$ fixed on block $B$ attached to the free end of spring $S_{B}$ detects this sound. The blocks $A$ and $B$ are simultaneously displaced towards each other through a distance of $1.0 m$ and then left to vibrate. If vibrational frequency of each block is $3 H z$ , the maximum frequencies of sound detected by $D$ is
$(V_{s o u n d} = 340 m / s)$

1

235 H z

2

564 H z

3

448 H z

4

668 H z

Explanation:

$ω = 2 π ν = 2 π 3 = 6 π$
$∴ ω = 18.85 {r a d s}^{- 1}$
Maximum speed of motion of $| v_{A} |$ and $| v_{B} | = A ω$
$f_{max} = (\frac{v + v_{B}}{v - v_{A}}) f = (\frac{340 + A ω}{340 - A ω}) f$
$= (\frac{340 + 18.85}{340 - 18.85}) 400$
$= (\frac{358.85}{321.15}) \times 400$
$= 1.12 \times 400$
$∴ f_{max} = 448 H z$

PHXI15:WAVES

354702 A bat moving at $10 m s^{- 1}$ towards a wall sends a sound signal of $8000 H z$ towards it. On refelection it hears a sound of frequency $f$ . The value of $f$ in $H z$ is close to (speed of sound $= 320 m s^{- 1}$ )

1 8516

2 8258

3 8424

4 8000

Explanation:

Reflected frequency of sound reaching bat
$= [\frac{v - (- v_{0})}{v - v_{s}}] f = [\frac{v + v_{0}}{v - v_{s}}] f = \frac{v + 10}{v - 10} f$
$= (\frac{320 + 10}{320 - 10}) \times 8000 = 8516 H z$

JEE - 2014

PHXI15:WAVES

354703 A boy is walking away from a wall at a speed of $1.0 m / s$ as shown in the figure. The boy blows a whistle steadily. An observer towards whom the boy is moving hears 4 beats. If the speed of sound is $340 m / s$ , the frequency of whistle is:
supporting img

1

480 H z

2

680 H z

3

840 H z

4

100 H z

Explanation:

Direct frequency received by the observer is
$f_{1} = f (\frac{340}{340 - 1}) (1)$
Frequency received by the wall is
$f_{2} = f (\frac{340}{340 + 1}) (2)$
Frequency received by the observer after reflection from the wall is
$f_{3} = f_{2}$
Given that
$f_{1} \sim f_{3} = 4$
$f [\frac{340}{339} - \frac{340}{341}] = 4$
$\Rightarrow f = 680 H z$

PHXI15:WAVES

354704 A person with vibrating tuning fork of frequency $338 H z$ is moving towards a vertical wall with a speed of $2 m s^{- 1}$ . Velocity of sound in air is 340 $m s^{- 1}$ . The number of beats heard by that person per second is

1 2

2 4

3 6

4 8

Explanation:

Frequency of tuning fork (source), $f = 338 H z$
Velocity of sound, $v = 340 m s^{- 1}$
When the person is moving towards the wall, the frequency of sound reaching the wall is
$f^{'} = \frac{f v}{v - v_{S}} (1)$
Where $v_{s} =$ speed of the person $= 2 m s^{- 1}$ .
On reflection the wall acts as source and the person is the listener moving towards the source. So the frequency heard by the person after reflection from the wall is
$f^{''} = [\frac{v + v_{L}}{v}] f^{'} (2)$
Substituting the value of $f^{'}$ from eq. (1) in (2), we get
$f^{''} = [\frac{v + v_{L}}{v - v_{S}}] f = (\frac{v + v_{S}}{v - v_{S}}) f [A s v_{L} = v_{S}]$
$= [\frac{340 + 2}{340 - 2}] \times 338 = 342 H z$
$∴$ Number of beats heard per second $f^{''} - f = 342 - 338 = 4$

KCET - 2013

PHXI15:WAVES

354705 supporting img
A source s emitting sound of $400 H z$ is fixed on block $A$ which is attached to free end of a spring $S_{A}$ . The detector $D$ fixed on block $B$ attached to the free end of spring $S_{B}$ detects this sound. The blocks $A$ and $B$ are simultaneously displaced towards each other through a distance of $1.0 m$ and then left to vibrate. If vibrational frequency of each block is $3 H z$ , the maximum frequencies of sound detected by $D$ is
$(V_{s o u n d} = 340 m / s)$

1

235 H z

2

564 H z

3

448 H z

4

668 H z

Explanation:

$ω = 2 π ν = 2 π 3 = 6 π$
$∴ ω = 18.85 {r a d s}^{- 1}$
Maximum speed of motion of $| v_{A} |$ and $| v_{B} | = A ω$
$f_{max} = (\frac{v + v_{B}}{v - v_{A}}) f = (\frac{340 + A ω}{340 - A ω}) f$
$= (\frac{340 + 18.85}{340 - 18.85}) 400$
$= (\frac{358.85}{321.15}) \times 400$
$= 1.12 \times 400$
$∴ f_{max} = 448 H z$

PHXI15:WAVES

354702 A bat moving at $10 m s^{- 1}$ towards a wall sends a sound signal of $8000 H z$ towards it. On refelection it hears a sound of frequency $f$ . The value of $f$ in $H z$ is close to (speed of sound $= 320 m s^{- 1}$ )

1 8516

2 8258

3 8424

4 8000

Explanation:

Reflected frequency of sound reaching bat
$= [\frac{v - (- v_{0})}{v - v_{s}}] f = [\frac{v + v_{0}}{v - v_{s}}] f = \frac{v + 10}{v - 10} f$
$= (\frac{320 + 10}{320 - 10}) \times 8000 = 8516 H z$

JEE - 2014

PHXI15:WAVES

354703 A boy is walking away from a wall at a speed of $1.0 m / s$ as shown in the figure. The boy blows a whistle steadily. An observer towards whom the boy is moving hears 4 beats. If the speed of sound is $340 m / s$ , the frequency of whistle is:
supporting img

1

480 H z

2

680 H z

3

840 H z

4

100 H z

Explanation:

Direct frequency received by the observer is
$f_{1} = f (\frac{340}{340 - 1}) (1)$
Frequency received by the wall is
$f_{2} = f (\frac{340}{340 + 1}) (2)$
Frequency received by the observer after reflection from the wall is
$f_{3} = f_{2}$
Given that
$f_{1} \sim f_{3} = 4$
$f [\frac{340}{339} - \frac{340}{341}] = 4$
$\Rightarrow f = 680 H z$

PHXI15:WAVES

354704 A person with vibrating tuning fork of frequency $338 H z$ is moving towards a vertical wall with a speed of $2 m s^{- 1}$ . Velocity of sound in air is 340 $m s^{- 1}$ . The number of beats heard by that person per second is

1 2

2 4

3 6

4 8

Explanation:

Frequency of tuning fork (source), $f = 338 H z$
Velocity of sound, $v = 340 m s^{- 1}$
When the person is moving towards the wall, the frequency of sound reaching the wall is
$f^{'} = \frac{f v}{v - v_{S}} (1)$
Where $v_{s} =$ speed of the person $= 2 m s^{- 1}$ .
On reflection the wall acts as source and the person is the listener moving towards the source. So the frequency heard by the person after reflection from the wall is
$f^{''} = [\frac{v + v_{L}}{v}] f^{'} (2)$
Substituting the value of $f^{'}$ from eq. (1) in (2), we get
$f^{''} = [\frac{v + v_{L}}{v - v_{S}}] f = (\frac{v + v_{S}}{v - v_{S}}) f [A s v_{L} = v_{S}]$
$= [\frac{340 + 2}{340 - 2}] \times 338 = 342 H z$
$∴$ Number of beats heard per second $f^{''} - f = 342 - 338 = 4$

KCET - 2013

PHXI15:WAVES

354705 supporting img
A source s emitting sound of $400 H z$ is fixed on block $A$ which is attached to free end of a spring $S_{A}$ . The detector $D$ fixed on block $B$ attached to the free end of spring $S_{B}$ detects this sound. The blocks $A$ and $B$ are simultaneously displaced towards each other through a distance of $1.0 m$ and then left to vibrate. If vibrational frequency of each block is $3 H z$ , the maximum frequencies of sound detected by $D$ is
$(V_{s o u n d} = 340 m / s)$

1

235 H z

2

564 H z

3

448 H z

4

668 H z

Explanation:

$ω = 2 π ν = 2 π 3 = 6 π$
$∴ ω = 18.85 {r a d s}^{- 1}$
Maximum speed of motion of $| v_{A} |$ and $| v_{B} | = A ω$
$f_{max} = (\frac{v + v_{B}}{v - v_{A}}) f = (\frac{340 + A ω}{340 - A ω}) f$
$= (\frac{340 + 18.85}{340 - 18.85}) 400$
$= (\frac{358.85}{321.15}) \times 400$
$= 1.12 \times 400$
$∴ f_{max} = 448 H z$

PHXI15:WAVES

354702 A bat moving at $10 m s^{- 1}$ towards a wall sends a sound signal of $8000 H z$ towards it. On refelection it hears a sound of frequency $f$ . The value of $f$ in $H z$ is close to (speed of sound $= 320 m s^{- 1}$ )

1 8516

2 8258

3 8424

4 8000

Explanation:

Reflected frequency of sound reaching bat
$= [\frac{v - (- v_{0})}{v - v_{s}}] f = [\frac{v + v_{0}}{v - v_{s}}] f = \frac{v + 10}{v - 10} f$
$= (\frac{320 + 10}{320 - 10}) \times 8000 = 8516 H z$

JEE - 2014

PHXI15:WAVES

354703 A boy is walking away from a wall at a speed of $1.0 m / s$ as shown in the figure. The boy blows a whistle steadily. An observer towards whom the boy is moving hears 4 beats. If the speed of sound is $340 m / s$ , the frequency of whistle is:
supporting img

1

480 H z

2

680 H z

3

840 H z

4

100 H z

Explanation:

Direct frequency received by the observer is
$f_{1} = f (\frac{340}{340 - 1}) (1)$
Frequency received by the wall is
$f_{2} = f (\frac{340}{340 + 1}) (2)$
Frequency received by the observer after reflection from the wall is
$f_{3} = f_{2}$
Given that
$f_{1} \sim f_{3} = 4$
$f [\frac{340}{339} - \frac{340}{341}] = 4$
$\Rightarrow f = 680 H z$

PHXI15:WAVES

354704 A person with vibrating tuning fork of frequency $338 H z$ is moving towards a vertical wall with a speed of $2 m s^{- 1}$ . Velocity of sound in air is 340 $m s^{- 1}$ . The number of beats heard by that person per second is

1 2

2 4

3 6

4 8

Explanation:

Frequency of tuning fork (source), $f = 338 H z$
Velocity of sound, $v = 340 m s^{- 1}$
When the person is moving towards the wall, the frequency of sound reaching the wall is
$f^{'} = \frac{f v}{v - v_{S}} (1)$
Where $v_{s} =$ speed of the person $= 2 m s^{- 1}$ .
On reflection the wall acts as source and the person is the listener moving towards the source. So the frequency heard by the person after reflection from the wall is
$f^{''} = [\frac{v + v_{L}}{v}] f^{'} (2)$
Substituting the value of $f^{'}$ from eq. (1) in (2), we get
$f^{''} = [\frac{v + v_{L}}{v - v_{S}}] f = (\frac{v + v_{S}}{v - v_{S}}) f [A s v_{L} = v_{S}]$
$= [\frac{340 + 2}{340 - 2}] \times 338 = 342 H z$
$∴$ Number of beats heard per second $f^{''} - f = 342 - 338 = 4$

KCET - 2013

PHXI15:WAVES

354705 supporting img
A source s emitting sound of $400 H z$ is fixed on block $A$ which is attached to free end of a spring $S_{A}$ . The detector $D$ fixed on block $B$ attached to the free end of spring $S_{B}$ detects this sound. The blocks $A$ and $B$ are simultaneously displaced towards each other through a distance of $1.0 m$ and then left to vibrate. If vibrational frequency of each block is $3 H z$ , the maximum frequencies of sound detected by $D$ is
$(V_{s o u n d} = 340 m / s)$

1

235 H z

2

564 H z

3

448 H z

4

668 H z

Explanation:

$ω = 2 π ν = 2 π 3 = 6 π$
$∴ ω = 18.85 {r a d s}^{- 1}$
Maximum speed of motion of $| v_{A} |$ and $| v_{B} | = A ω$
$f_{max} = (\frac{v + v_{B}}{v - v_{A}}) f = (\frac{340 + A ω}{340 - A ω}) f$
$= (\frac{340 + 18.85}{340 - 18.85}) 400$
$= (\frac{358.85}{321.15}) \times 400$
$= 1.12 \times 400$
$∴ f_{max} = 448 H z$

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