An acoustic pulse generator located at a fixed point between two parallel rigid reflective… — JAMB UTME

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 $748$ wavelengthsAnswer
B
$748\text{ m}$ and $1496$ wavelengths
C
$102\text{ m}$ and $204$ wavelengths
D
$374\text{ m}$ and $187$ wavelengths

Answer

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}

\frac{374}{0.5} = 748

Step-by-Step Solution

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.

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.

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.

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.

Key Concept

Echo distance calculations and wave speed-frequency-wavelength relationships

Estimated Time:2m 0s

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