Sound Waves, Echoes, Pitch, Loudness, and Quality Alıştırma Soruları — JAMB UTME

Soru 1Soru →

A ship uses a sonar device to determine the depth of the ocean floor. A sound pulse sent vertically downward reflects off the seabed and is detected $0.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at that location?

450\text{ m}

900\text{ m}

1800\text{ m}

2500\text{ m}

Cevabı ve açıklamayı göster

Cevap: $450\text{ m}$

Cevap

450\text{ m}

The correct answer is

450\text{ m}

because sound travels to the ocean floor and back, covering twice the actual depth. Dividing the total distance of

900\text{ m}

by

2

yields the one-way depth of

450\text{ m}

.

Adım Adım Çözüm

1

Calculate the total distance traveled by the sound wave

d_{\text{total}} = 1500\text{ m/s} \times 0.6\text{ s} = 900\text{ m}

Total distance traveled by the wave equals speed multiplied by total elapsed time.

2

Determine the one-way depth of the ocean floor

\text{Depth} = \frac{d_{\text{total}}}{2} = \frac{900\text{ m}}{2} = 450\text{ m}

An echo travels down to the seabed and back to the receiver, so the ocean depth is half the total distance covered.

Anahtar Kavram

Echo distance calculation

Tahmini Süre:45s

Soru 2Soru →

A man standing at a distance in front of a tall vertical cliff claps his hands and hears the echo after $1.2\text{ s}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the man and the cliff in meters?

Cevabı ve açıklamayı göster

Cevap: 204

Cevap

The distance between the man and the cliff is

204\text{ m}

.

An echo is formed when sound travels to an obstacle and reflects back. The time taken for the sound to travel to the cliff and back is

1.2\text{ s}

. The total distance covered by sound is

v \times t = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

. Since this distance covers two equal trips (to the cliff and back), the distance to the cliff is

408\text{ m} / 2 = 204\text{ m}

.

Adım Adım Çözüm

1

State the relationship between sound speed, total echo time, and distance.

Total distance traveled by the sound is twice the distance to the cliff:

2d = v \times t

.

An echo involves sound traveling from the source to the reflecting barrier and back.

2

Substitute the given values into the equation.

2d = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

.

To find the total distance traversed by the sound wave.

3

Solve for the distance

d

.

d = \frac{408\text{ m}}{2} = 204\text{ m}

.

The one-way distance to the cliff is half of the total distance traveled by the echo.

Anahtar Kavram

Calculation of echo distance using

d = \frac{v t}{2}

Soru 3Soru →

A student stands between two tall, parallel vertical walls and claps her hands once. She hears the first echo after $1.0\text{ s}$ and the second echo after $1.5\text{ s}$ . Taking the speed of sound in air to be $340\text{ m s}^{-1}$ , what is the distance between the two walls?

425\text{ m}

850\text{ m}

510\text{ m}

255\text{ m}

Cevabı ve açıklamayı göster

Cevap: $425\text{ m}$

Cevap

The distance between the two walls is

425\text{ m}

.

Sound travels from the student to each wall and reflects back. The distance to the first wall is

\frac{340 \times 1.0}{2} = 170\text{ m}

and to the second wall is

\frac{340 \times 1.5}{2} = 255\text{ m}

. Adding both distances gives the total separation between the walls as

425\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance from the student to the closer wall (

d_1

)

d_1 = \frac{v \times t_1}{2} = \frac{340 \text{ m s}^{-1} \times 1.0 \text{ s}}{2} = 170\text{ m}

Sound travels to the wall and reflects back, so the time given corresponds to twice the distance.

2

Calculate the distance from the student to the further wall (

d_2

)

d_2 = \frac{v \times t_2}{2} = \frac{340 \text{ m s}^{-1} \times 1.5 \text{ s}}{2} = 255\text{ m}

The sound for the second echo travels to the second wall and back.

3

Determine the total distance between the two parallel walls

D = d_1 + d_2 = 170\text{ m} + 255\text{ m} = 425\text{ m}

Since the student is positioned between the two walls, the separation of the walls is the sum of both individual distances.

Anahtar Kavram

Echo distance relation

2d = v t

for sound reflection from barriers.

Tahmini Süre:1m 30s

Soru 4Soru →

An observer standing at a position between two tall parallel vertical cliffs fires a signal pistol. The observer hears the first echo reflected from the nearer cliff after $1.2\text{ s}$ and the second echo reflected from the farther cliff after $1.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ , what is the total distance between the two cliffs?

102\text{ m}

510\text{ m}

816\text{ m}

1020\text{ m}

Cevabı ve açıklamayı göster

Cevap: $510\text{ m}$

Cevap

The total distance between the two cliffs is

510\text{ m}

.

Because an echo involves two-way travel of sound, the distance

d

from an observer to a reflecting barrier is given by

d = \frac{v t}{2}

. For the nearer cliff, the distance is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

. For the farther cliff, the distance is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. The total distance between the parallel cliffs is

d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance from the observer to the nearer cliff

d_1 = \frac{v \times t_1}{2} = \frac{340\text{ m s}^{-1} \times 1.2\text{ s}}{2} = 204\text{ m}

An echo involves sound traveling from the observer to the cliff and back, so the one-way distance is half the total path length.

2

Calculate the distance from the observer to the farther cliff

d_2 = \frac{v \times t_2}{2} = \frac{340\text{ m s}^{-1} \times 1.8\text{ s}}{2} = 306\text{ m}

Similarly, the sound travels to the farther cliff and back in

1.8\text{ s}

.

3

Sum the two one-way distances to find the total distance between the two cliffs

D = d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

Since the observer is positioned between the two parallel cliffs, the total separation distance is the sum of the individual distances to each cliff.

Anahtar Kavram

Echo distance calculation involving two-way sound propagation

Soru 5Soru →

An ultrasonic rangefinder mounted on a drone sends a sound pulse vertically downward to measure its altitude above flat ground. If the echo is detected by the sensor $0.08\text{ s}$ after emission and the speed of sound in air is $340\text{ m s}^{-1}$ , what is the altitude of the drone in meters?

Cevabı ve açıklamayı göster

Cevap: 13.6

Cevap

13.6 meters

The sound pulse emitted by the drone travels down to the ground and reflects back to the sensor. The relationship between speed

v

, total round-trip time

t

, and altitude

d

is given by

2d = v \times t

. Substituting

v = 340\text{ m s}^{-1}

and

t = 0.08\text{ s}

yields

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Adım Adım Çözüm

1

Identify the total time taken by the sound pulse for the round trip.

Total round-trip time

t = 0.08\text{ s}

and speed of sound

v = 340\text{ m s}^{-1}

.

Echo detection measures the time for sound to travel to a barrier and return.

2

Apply the echo distance relationship

2d = v \times t

to solve for altitude

d

.

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Dividing the total path distance by 2 yields the one-way distance to the ground.

Anahtar Kavram

Calculation of distance using echoes and two-way sound wave propagation

Alternatif Yöntem

Determine the one-way travel time first:

t_{\text{one-way}} = \frac{0.08}{2} = 0.04\text{ s}

. Then calculate altitude directly using distance = speed × one-way time:

d = 340 \times 0.04 = 13.6\text{ m}

.

Tahmini Süre:1m 0s

Soru 6Soru →

A motorist travelling at a constant speed of $20\text{ m s}^{-1}$ directly towards a tall vertical cliff sounds a horn. If the motorist hears the echo of the horn $2.0\text{ s}$ later and the speed of sound in air is $340\text{ m s}^{-1}$ , what was the distance of the car from the cliff at the moment the horn was sounded?

Cevabı ve açıklamayı göster

Cevap: 360

Cevap

The distance of the car from the cliff at the instant the horn was sounded was

360\text{ m}

.

When the motorist sounds the horn at an initial distance

D

from the cliff, the sound wave travels toward the cliff. In the

2.0\text{ s}

it takes for the echo to return, the car advances

40\text{ m}

toward the cliff (

20\text{ m s}^{-1} \times 2.0\text{ s}

). The returning echo meets the motorist at a distance of

(D - 40)\text{ m}

from the cliff. Consequently, the sound covers a total distance of

D + (D - 40) = 2D - 40\text{ m}

. Because the sound wave travels at

340\text{ m s}^{-1}

for

2.0\text{ s}

, the actual distance covered by sound is

340 \times 2.0 = 680\text{ m}

. Setting

2D - 40 = 680

gives

2D = 720\text{ m}

, which yields

D = 360\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance covered by the car while moving toward the cliff during the echo time interval.

d_{\text{car}} = 20\text{ m s}^{-1} \times 2.0\text{ s} = 40\text{ m}

.

The car continues to move closer to the cliff for the entire

2.0\text{ s}

period.

2

Set up an expression for the total distance covered by the sound wave.

d_{\text{sound}} = D + (D - 40) = 2D - 40\text{ m}

.

The sound travels forward a distance

D

to the cliff and reflects back to the car's updated location, which is

(D - 40)\text{ m}

from the cliff.

3

Calculate the distance travelled by sound using the given speed of sound.

d_{\text{sound}} = 340\text{ m s}^{-1} \times 2.0\text{ s} = 680\text{ m}

.

Sound propagates through air at

340\text{ m s}^{-1}

.

4

Equate the geometric path expression to the physical sound distance and solve for

D

.

2D - 40 = 680 \implies 2D = 720 \implies D = 360\text{ m}

.

Solving the equation yields the initial position of the car relative to the cliff.

Anahtar Kavram

Echo distance calculation with a moving observer

Soru 7Soru →

A train moving at a constant speed of $15\text{ m/s}$ towards a tall vertical cliff emits a whistle signal. The driver hears the echo of the whistle $2.0\text{ s}$ after it was sounded. If the speed of sound in air is $340\text{ m/s}$ , what was the distance between the train and the cliff at the instant the whistle was sounded?

355\text{ m}

340\text{ m}

710\text{ m}

325\text{ m}

Cevabı ve açıklamayı göster

Cevap: $355\text{ m}$

Cevap

The initial distance between the train and the cliff when the whistle was sounded was

355\text{ m}

.

In

2.0\text{ s}

, the sound wave travels

680\text{ m}

while the train advances

30\text{ m}

towards the cliff. The total path of the sound consists of the initial distance

D

to the cliff plus the return distance

(D - 30)\text{ m}

to the moving train. Equating

D + (D - 30) = 680\text{ m}

gives

2D = 710\text{ m}

, which yields an initial distance of

355\text{ m}

.

Adım Adım Çözüm

1

Calculate the total distance traveled by the sound wave in the given time interval.

s_{\text{sound}} = v_{\text{sound}} \times t = 340\text{ m/s} \times 2.0\text{ s} = 680\text{ m}

.

Sound travels at a constant speed in air over the total elapsed time of

2.0\text{ s}

.

2

Calculate the distance covered by the moving train during the same time interval.

s_{\text{train}} = v_{\text{train}} \times t = 15\text{ m/s} \times 2.0\text{ s} = 30\text{ m}

.

The train continues moving towards the cliff while the sound wave travels to the cliff and reflects back.

3

Formulate the geometric path equation for the sound wave.

Let

D

be the initial distance to the cliff. Sound travels

D

to the cliff and reflects back a distance of

(D - 30)\text{ m}

to reach the train. Therefore,

D + (D - 30) = 680\text{ m}

.

The echo is received by the driver at a position

30\text{ m}

closer to the cliff than where the sound was emitted.

4

Solve the linear equation for

D

.

2D - 30 = 680 \implies 2D = 710 \implies D = 355\text{ m}

.

Solving for

D

gives the exact distance at the moment the whistle was sounded.

Anahtar Kavram

Echo path geometry with a moving sound source

Soru 8Soru →

An acoustic pulse generator located at a fixed point between two parallel rigid reflective barriers emits a sound wave with a frequency of $680\text{ Hz}$ . The first echo from the closer barrier is detected after $0.8\text{ s}$ , and the first echo from the farther barrier is detected after $1.4\text{ s}$ . Assuming the speed of sound in the medium is $340\text{ m s}^{-1}$ , what is the total distance between the two barriers, and how many complete wavelengths of this sound wave fit within this total distance?

374\text{ m}

and

748

wavelengths

748\text{ m}

and

1496

wavelengths

102\text{ m}

and

204

wavelengths

374\text{ m}

and

187

wavelengths

Cevabı ve açıklamayı göster

Cevap: $374\text{ m}$ and $748$ wavelengths

Cevap

The total distance between the barriers is

374\text{ m}

and

748

complete wavelengths fit within this distance.

The distance to the first barrier is

d_1 = \frac{340 \times 0.8}{2} = 136\text{ m}

, and to the second barrier is

d_2 = \frac{340 \times 1.4}{2} = 238\text{ m}

. Adding both distances gives a total separation of

374\text{ m}

. With a wavelength

\lambda = \frac{v}{f} = \frac{340}{680} = 0.5\text{ m}

, the number of full wavelengths fitting in

374\text{ m}

is

\frac{374}{0.5} = 748

.

Adım Adım Çözüm

1

Calculate the one-way distance from the generator to each reflective barrier using the echo relationship

d = \frac{v \cdot t}{2}

.

For barrier 1:

d_1 = \frac{340 \text{ m s}^{-1} \times 0.8 \text{ s}}{2} = 136 \text{ m}

. For barrier 2:

d_2 = \frac{340 \text{ m s}^{-1} \times 1.4 \text{ s}}{2} = 238 \text{ m}

.

An echo represents a two-way journey (to the surface and back), so the time taken to travel the one-way distance is half of the echo reception time.

2

Determine the total distance between the two parallel barriers.

D = d_1 + d_2 = 136 \text{ m} + 238 \text{ m} = 374 \text{ m}

.

Since the generator is situated between the two barriers, the total separation distance equals the sum of the individual distances to each barrier.

3

Calculate the wavelength

\lambda

of the sound wave using the wave equation

v = f \lambda

.

\lambda = \frac{v}{f} = \frac{340 \text{ m s}^{-1}}{680 \text{ Hz}} = 0.5 \text{ m}$.

Wavelength is the ratio of wave speed to frequency.

4

Compute the number of complete wavelengths

N

contained within the total separation distance.

N = \frac{D}{\lambda} = \frac{374 \text{ m}}{0.5 \text{ m}} = 748.

Dividing total distance by single wavelength yields the number of full wave cycles occupying that span.

Anahtar Kavram

Echo distance calculations and wave speed-frequency-wavelength relationships

Tahmini Süre:2m 0s

Soru 9Soru →

A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end records two sound signals—one traveling through the aluminum pipeline and the other through the surrounding air—separated by a time interval of $2.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ and the speed of sound in aluminum is $5100\text{ m s}^{-1}$ , what is the length of the pipeline in meters?

Cevabı ve açıklamayı göster

Cevap: 1020

Cevap

The length of the pipeline is

1020\text{ m}

.

Sound travels significantly faster through solids like aluminum (

5100\text{ m s}^{-1}

) than through gases like air (

340\text{ m s}^{-1}

). The time taken for sound to travel a distance

L

through air is

t_{\text{air}} = \frac{L}{340}

, while through aluminum it is

t_{\text{metal}} = \frac{L}{5100}

. Setting their difference equal to

2.8\text{ s}

gives

\frac{L}{340} - \frac{L}{5100} = 2.8

, which solves to

L = 1020\text{ m}

.

Adım Adım Çözüm

1

Formulate transit time expressions for both media

t_{\text{air}} = \frac{L}{340}

and

t_{\text{metal}} = \frac{L}{5100}

Time taken by a wave to travel distance

L

at constant speed

v

is

t = \frac{L}{v}

.

2

Set up the time difference equation

t_{\text{air}} - t_{\text{metal}} = 2.8\text{ s}

The sound wave travels faster through aluminum than air, so the air pulse arrives later by

2.8\text{ s}

.

3

Solve the algebraic equation for distance

L

L = 1020\text{ m}

Combining terms yields

\frac{14L}{5100} = 2.8

, which simplifies to

L = \frac{2.8 \times 5100}{14} = 1020\text{ m}

.

Anahtar Kavram

Propagation speed of sound waves in different physical media

Soru 10Soru →

An echo sounder on a fishing boat emits an acoustic pulse vertically downward toward the seabed and detects the reflected signal $0.80\text{ s}$ later. If the speed of sound in seawater is $1450\text{ m/s}$ , what is the depth of the sea in meters at this location?

Cevabı ve açıklamayı göster

Cevap: 580

Cevap

The depth of the sea is

580\text{ m}

.

An echo signal travels to the reflecting surface and back, covering twice the depth (

2d = v \times t

). Substituting

v = 1450\text{ m/s}

and

t = 0.80\text{ s}

gives

2d = 1160\text{ m}

, so the seabed depth

d = 580\text{ m}

.

Adım Adım Çözüm

1

Identify the given physical quantities

Total travel time

t = 0.80\text{ s}

and speed of sound

v = 1450\text{ m/s}

.

An echo involves sound traveling to the seabed and back, so the recorded time represents a two-way journey.

2

Set up the distance equation for an echo

Total distance traveled by the sound pulse is

2d = v \times t

.

Sound travels to the sea floor and reflects back to the ship, covering a total distance equal to twice the depth.

3

Calculate the depth

d

d = \frac{1450 \times 0.80}{2} = 580\text{ m}

.

Dividing the total round-trip distance by 2 yields the one-way depth.

Anahtar Kavram

Calculation of distance using sound echoes in a medium

Soru 11Soru →

A girl stands at a specific distance from a flat vertical wall and claps her hands once. If she hears the echo $0.4\text{ s}$ later, what is her distance from the wall in meters? (Take the speed of sound in air as $340\text{ m/s}$ )

Cevabı ve açıklamayı göster

Cevap: 68

Cevap

The distance of the girl from the wall is

68\text{ m}

.

An echo involves the sound traveling to the reflecting surface and returning to the source, covering a total distance of

2d

. Using

2d = v \times t

, we obtain

d = \frac{340 \times 0.4}{2} = 68\text{ m}

.

Adım Adım Çözüm

1

Identify the given values from the problem statement.

Time for echo

t = 0.4\text{ s}

, speed of sound

v = 340\text{ m/s}

.

An echo is a reflected sound wave that travels to the wall and back, completing a round trip.

2

Apply the echo calculation formula.

d = \frac{v \times t}{2}

The total distance traveled by the sound is

2d

. Thus, the one-way distance

d

to the reflecting surface is half of the total distance.

3

Substitute the values and calculate the distance.

d = \frac{340 \times 0.4}{2} = 68\text{ m}

Multiplying the speed by half the elapsed time gives the distance to the wall.

Anahtar Kavram

Echo distance calculation

Soru 12Soru →

A sound transmitter and a projectile launcher are co-located at a distance of $210\text{ m}$ directly in front of a tall, flat vertical cliff. At time $t = 0\text{ s}$ , a projectile is launched directly away from the cliff at a constant speed of $60\text{ m s}^{-1}$ , while a sound pulse is emitted simultaneously towards the cliff. The sound wave reflects off the cliff face and travels back to overtake the moving projectile. Assuming the speed of sound in air is $340\text{ m s}^{-1}$ , calculate the distance from the cliff face, in meters, to the position where the reflected sound wave intercepts the projectile.

Cevabı ve açıklamayı göster

Cevap: 300

Cevap

The distance of the projectile from the cliff face at the instant of interception is

300\text{ m}

.

The correct calculation accounts for both the two-part path of the sound wave (forward to cliff + back to projectile) and the displacement of the projectile moving away from the cliff over the same time interval, yielding an interception distance of

300\text{ m}

from the cliff.

Adım Adım Çözüm

1

Set up the total distance expression for the sound wave from the cliff face

Total sound path =

210\text{ m} + x

The sound pulse must travel

210\text{ m}

forward to hit the cliff face, plus an additional distance

x

away from the cliff face after reflection to reach the projectile.

2

Set up the distance expression for the projectile from its starting point

Projectile path =

x - 210\text{ m}

The projectile starts

210\text{ m}

away from the cliff and moves farther away to position

x

.

3

Equate the time elapsed for both sound propagation and projectile movement

\frac{210 + x}{340} = \frac{x - 210}{60}

Both events happen simultaneously over the exact same time interval

t

.

4

Solve the linear equation for

x

x = 300\text{ m}

Cross-multiplying yields

60(210 + x) = 340(x - 210)

, which simplifies to

28x = 8400

, giving

x = 300\text{ m}

.

Anahtar Kavram

Echo reflection path combined with relative linear kinematics

Soru 13Soru →

A survey ship emits an ultrasonic sound pulse vertically downwards into the ocean. The sound wave travels through seawater at a speed of $1500\text{ m/s}$ , and its reflected echo from the seabed is detected by the ship's hydrophone $1.2\text{ s}$ after emission. What is the depth of the ocean floor at that location?

900\text{ m}

1800\text{ m}

450\text{ m}

2700\text{ m}

Cevabı ve açıklamayı göster

Cevap: $900\text{ m}$

Cevap

900\text{ m}

Sound emitted by the ship travels down to the seabed and reflects back to the hydrophone, taking

1.2\text{ s}

for the complete round trip. Using the relation

\text{Depth} = \frac{v \times t}{2}

, the depth is

\frac{1500 \times 1.2}{2} = 900\text{ m}

.

Adım Adım Çözüm

1

Calculate total distance traveled by the sound wave

Total distance

s = v \times t = 1500\text{ m/s} \times 1.2\text{ s} = 1800\text{ m}

Distance is the product of speed and total time elapsed.

2

Determine the depth of the seabed

Depth

d = \frac{s}{2} = \frac{1800\text{ m}}{2} = 900\text{ m}

An echo involves the sound traveling down to the ocean floor and back up, so the depth is half the total distance.

Anahtar Kavram

Echo Reflection and Distance Calculation

Tahmini Süre:45s

Soru 14Soru →

A research vessel emits a high-frequency acoustic pulse vertically downward toward the seabed. The signal reflects off the ocean floor and is detected by the vessel's receiver $1.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at this point?

2400\text{ m}

1200\text{ m}

600\text{ m}

937.5\text{ m}

Cevabı ve açıklamayı göster

Cevap: $1200\text{ m}$

Cevap

The depth of the ocean floor is

1200\text{ m}

.

The correct answer is

1200\text{ m}

. Since an echo involves a two-way journey (from the vessel down to the ocean floor and back up), the sound wave takes half of the total time (

0.8\text{ s}

) to reach the bottom. Multiplying the speed of sound in seawater (

1500\text{ m/s}

) by

0.8\text{ s}

yields the correct depth of

1200\text{ m}

.

Adım Adım Çözüm

1

Determine the time taken for the sound wave to travel one way to the ocean floor.

t_{\text{one-way}} = \frac{t_{\text{total}}}{2} = \frac{1.6\text{ s}}{2} = 0.8\text{ s}

An echo involves sound traveling from the transmitter to the reflecting surface and back to the receiver.

2

Calculate the depth using the speed of sound in seawater.

\text{Depth } d = v \times t_{\text{one-way}} = 1500\text{ m/s} \times 0.8\text{ s} = 1200\text{ m}

The distance traveled in one direction equals the speed of sound multiplied by the one-way travel time.

Anahtar Kavram

Echo location and depth sounding using two-way wave propagation

Tahmini Süre:1m 0s

Soru 15Soru →

A research submarine moving underwater at a constant speed of $12.0\text{ m/s}$ directly toward a vertical underwater cliff face emits an ultrasonic acoustic pulse. The echo reflected from the cliff face is detected by the submarine's receiver $2.50\text{ s}$ after emission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the distance between the submarine and the cliff face at the exact moment the echo is detected?

Cevabı ve açıklamayı göster

Cevap: 1860

Cevap

The distance between the submarine and the cliff face at the exact moment the echo is detected is

1860\text{ m}

.

During the

2.50\text{ s}

transit time of the acoustic signal, the sound covers a total path of

3750\text{ m}

(

1500\text{ m/s} \times 2.50\text{ s}

) while the submarine moves

30\text{ m}

closer to the cliff face (

12.0\text{ m/s} \times 2.50\text{ s}

). The total path of the sound consists of the outward journey to the cliff (

d + 30\text{ m}

) and the return journey to the submarine (

d

). Setting

(d + 30) + d = 3750

gives

2d + 30 = 3750

, leading to

d = 1860\text{ m}

.

Adım Adım Çözüm

1

Calculate total sound travel distance and submarine displacement during the 2.50 s window.

Sound distance

d_{\text{sound}} = 1500\text{ m/s} \times 2.50\text{ s} = 3750\text{ m}

; Submarine displacement

d_{\text{sub}} = 12.0\text{ m/s} \times 2.50\text{ s} = 30.0\text{ m}

.

Both the acoustic wave and the submarine move continuously throughout the total elapsed transit time.

2

Establish the geometric equation for the sound path relative to the final distance d.

d_{\text{sound}} = 2d + d_{\text{sub}}

, where

d

is the remaining distance to the cliff face at detection time.

The sound pulse travels forward across the initial separation

(d + d_{\text{sub}})

and reflects back across the remaining separation

d

.

3

Solve the linear equation for the final separation distance d.

3750 = 2d + 30 \implies 2d = 3720 \implies d = 1860\text{ m}

.

Subtracting the submarine's forward displacement from the total sound path gives twice the distance to the obstacle at the instant of signal reception.

Anahtar Kavram

Echo distance calculations with moving receiver and source

Soru 16Soru →

A person standing at a stationary position between two tall parallel vertical walls fires a starter pistol. The person hears the first echo reflected from the nearer wall after $1.2\text{ s}$ and the second echo from the farther wall after $1.8\text{ s}$ . Given that the speed of sound in air is $340\text{ m/s}$ , what is the total distance between the two walls in meters?

Cevabı ve açıklamayı göster

Cevap: 510

Cevap

The total distance between the two walls is

510\text{ m}

.

Because sound travels to each wall and reflects back to the observer, the distance to each wall is given by

d = \frac{v \cdot t}{2}

. The distance to the nearer wall is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

, and the distance to the farther wall is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. Since the observer is between the two walls, the total separation between the walls is

204\text{ m} + 306\text{ m} = 510\text{ m}

.

Adım Adım Çözüm

1

Calculate the distance from the observer to the nearer wall.

d_1 = 204\text{ m}

The sound travels to the nearer wall and back in

1.2\text{ s}

, covering twice the distance to that wall.

2

Calculate the distance from the observer to the farther wall.

d_2 = 306\text{ m}

The sound travels to the farther wall and back in

1.8\text{ s}

, covering twice the distance to that wall.

3

Add the two individual distances to find the total distance between the walls.

D = d_1 + d_2 = 510\text{ m}

The observer is positioned between the two walls, so the separation distance is the sum of both distances.

Anahtar Kavram

Echo and Speed of Sound Propagation between Parallel Boundaries

Soru 17Soru →

A hiker standing at a distance from a tall vertical cliff shouts and hears an echo $0.60\text{ s}$ later. If the speed of sound in air is $330\text{ m/s}$ , what is the distance between the hiker and the cliff?

198\text{ m}

99\text{ m}

550\text{ m}

49.5\text{ m}

Cevabı ve açıklamayı göster

Cevap: $99\text{ m}$

Cevap

The distance between the hiker and the cliff is

99\text{ m}

.

An echo is a reflected sound wave. The sound travels from the hiker to the cliff and back, covering a total distance of

2d

in time

t

. Therefore,

2d = v \times t

, which gives

d = \frac{330\text{ m/s} \times 0.60\text{ s}}{2} = 99\text{ m}

.

Adım Adım Çözüm

1

Identify given parameters and echo relation

Speed of sound

v = 330\text{ m/s}

, time elapsed

t = 0.60\text{ s}

. An echo involves sound traveling to the wall and back (

2d

).

The total distance traveled by the sound wave during time

t

is twice the distance to the reflecting surface.

2

Calculate the one-way distance

d = \frac{v \times t}{2} = \frac{330 \times 0.60}{2} = 99\text{ m}

.

Dividing the total distance by 2 yields the actual distance from the hiker to the cliff.

Anahtar Kavram

Echo distance calculation

Soru 18Soru →

A musical note played on an instrument produces a fundamental frequency of $440\text{ Hz}$ . Given that the speed of sound in air is $330\text{ m/s}$ , what is the wavelength, in meters, of the sound wave produced in air corresponding to its second overtone?

Cevabı ve açıklamayı göster

Cevap: 0.25

Cevap

0.25 m

The fundamental frequency (

f_1 = 440\text{ Hz}

) is the first harmonic. The second overtone is the third harmonic, which has a frequency of

3 \times 440\text{ Hz} = 1320\text{ Hz}

. Substituting this into the wave speed equation

\lambda = \frac{v}{f}

gives

\lambda = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

.

Adım Adım Çözüm

1

Determine the harmonic number for the second overtone

The second overtone is the third harmonic (

n = 3

)

Overtones are integer harmonics above the fundamental frequency (

n = 1

). Thus, the 1st overtone is

n = 2

and the 2nd overtone is

n = 3

.

2

Calculate the frequency of the second overtone

f_3 = 3 \times 440\text{ Hz} = 1320\text{ Hz}

The frequency of the

n

-th harmonic is

n

times the fundamental frequency.

3

Calculate the wavelength using the wave speed equation

\lambda = \frac{v}{f_3} = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

Wavelength is determined by dividing the speed of sound by the wave frequency.

Anahtar Kavram

Harmonics, Overtones, and Wave Equation

Soru 19Soru →

A surveyor standing $255\text{ m}$ away from the base of a vertical canyon wall emits a short acoustic signal. If the speed of sound in air is $340\text{ m/s}$ , after what time interval will the surveyor detect the reflected echo?

0.75\text{ s}

1.50\text{ s}

1.33\text{ s}

3.00\text{ s}

Cevabı ve açıklamayı göster

Cevap: $1.50\text{ s}$

Cevap

1.50\text{ s}

Sound must travel to the cliff face and reflect back to the surveyor, covering a total distance of

2 \times 255\text{ m} = 510\text{ m}

. Using the speed formula

t = \frac{d}{v}

, the elapsed time is

\frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}

.

Adım Adım Çözüm

1

Determine the total distance traveled by the sound wave

s_{total} = 2 \times 255\text{ m} = 510\text{ m}

An echo requires sound to travel to the obstacle and reflect back to the source.

2

Calculate the time taken using the wave speed formula

t = \frac{s_{total}}{v} = \frac{510\text{ m}}{340\text{ m/s}} = 1.50\text{ s}

Time is equal to total distance divided by the speed of propagation.

Anahtar Kavram

Echo reflection and two-way sound propagation distance

Tahmini Süre:1m 0s

Soru 20Soru →

A bat emitting an ultrasonic sound pulse towards a flat vertical wall receives the reflected echo $0.12\text{ s}$ after emission. If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the bat and the wall at the moment of emission?

20.4\text{ m}

40.8\text{ m}

81.6\text{ m}

10.2\text{ m}

Cevabı ve açıklamayı göster

Cevap: $20.4\text{ m}$

Cevap

The distance between the bat and the wall is

20.4\text{ m}

.

The distance to the wall is

20.4\text{ m}

because sound undergoes a two-way journey (from source to reflector and back). Multiplying the speed of sound (

340\text{ m/s}

) by the total time (

0.12\text{ s}

) yields the round-trip distance of

40.8\text{ m}

. Dividing this value by

2

gives the one-way distance of

20.4\text{ m}

.

Adım Adım Çözüm

1

Calculate the total distance traveled by the sound wave during the round trip.

d_{\text{total}} = v \times t = 340\text{ m/s} \times 0.12\text{ s} = 40.8\text{ m}

The sound wave travels from the bat to the obstacle and reflects back to the bat.

2

Divide the total round-trip distance by

2

to determine the one-way distance to the wall.

d = \frac{d_{\text{total}}}{2} = \frac{40.8\text{ m}}{2} = 20.4\text{ m}

An echo involves a two-way path, so the one-way distance is half the total distance covered by the wave.

Anahtar Kavram

Echo and Two-Way Distance Calculation

Tahmini Süre:1m 30s

Sound Waves, Echoes, Pitch, Loudness, and Quality

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Alternatif Yöntem

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm