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Chapter 16: Problem 2

(1) A sailor strikes the side of his ship just below the waterline.He hears the echo of the sound reflected from the ocean floor directly below 2.5 \(\mathrm{s}\) blater. How deep is the ocean at this point? Assume the speed of Sound in sea water is 1560 \(\mathrm{m} / \mathrm{s}\) (Table 1\()\) and does not vary significantly \(\quad\) with depth.

Short Answer

Expert verified
The ocean depth is 1950 meters.

Step by step solution

01

Understand the problem

We need to determine the depth of the ocean. The time given is for the sound wave to travel to the ocean floor and back, and we know the speed of sound in seawater is 1560 m/s.
02

Calculate the total distance traveled by the sound

The sound travels from the ship to the ocean floor and back. Therefore, the total travel time is 2.5 seconds, which accounts for both the downward and upward journeys.
03

Use the speed formula to find total distance

The formula to calculate distance is: \( \, \text{Distance} = \text{Speed} \times \text{Time} \,\). The speed of sound in seawater is 1560 m/s, and the total time is 2.5 s. Therefore, the total distance is calculated as \( \, \text{Distance} = 1560 \, \text{m/s} \times 2.5 \, \text{s} = 3900 \, \text{m} \,\).
04

Calculate the depth of the ocean

Since the total distance (3900 m) accounts for the round trip of sound, the actual depth of the ocean is half of this. Therefore, \( \, \text{Depth} = \frac{3900 \, \text{m}}{2} = 1950 \, \text{m} \,\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sound waves in seawater
Sound travels as waves, which are disturbances that move through a medium. In seawater, sound waves behave differently than in air due to water's unique properties. Seawater is denser and more elastic than air, providing better conditions for sound to travel. This means that sound waves can move quickly and over long distances through the ocean.
Because of
  • increased density
  • elastic properties of water
sound waves in seawater carry energy efficiently. This is why submarines, sonar devices, and marine animals like whales use sound for communication and navigation underwater.
Understanding sound in water helps us solve problems related to underwater acoustics, such as locating ocean floors or understanding marine life behavior.
Speed of sound
The speed of sound is how fast sound waves travel through a medium. In seawater, the speed of sound is typically around 1560 m/s, as seen from the given problem. This higher speed compared to air is due to the increased density of water and its elastic properties.
Several factors can influence the speed of sound in seawater, including:
  • Temperature: Warmer water increases the speed because particles move more quickly.
  • Salinity: More dissolved salts increase the density, slightly affecting the speed.
  • Pressure: Depth impacts pressure, which also affects speed.
Calculating sound speed in seawater is crucial for applications like sonar and submarine navigation, allowing for precise measurements and understanding of underwater environments.
Echo reflection
An echo is a sound that reflects off a surface and returns to its source. This concept is employed in the exercise to determine the ocean's depth. When sound waves hit the ocean floor, they bounce back, creating an echo that can be detected by the sailor.
The process of echo reflection involves:
  • Sound waves traveling from the source to a surface (like the ocean floor).
  • Reflection when waves hit the surface, creating a return wave or echo.
  • Detecting the reflected wave, which helps measure distance.
In the problem, the time taken for the echo to return is used to calculate the distance to the ocean floor. By knowing the speed of sound and the round-trip time, we can determine the depth, showcasing echo reflection's practical use in measuring distances underwater.

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Most popular questions from this chapter

A science museum has a display called a sewer pipe symphony. It consists of many plastic pipes of various lengths, which are open on both ends. (a) If the pipes have lengths of \(3.0 \mathrm{m}, 2.5 \mathrm{m}\) , \(2.0 \mathrm{m}, 1.5 \mathrm{m}\) and \(1.0 \mathrm{m},\) what frequencies will be heard by a visitor's ear placed near the ends of the pipes? (b) Why does this display work better on a noisy day than on a quiet day?

An airplane travels at Mach 2.0 where the speed of sound is \(310 \mathrm{~m} / \mathrm{s} .(a)\) What is the angle the shock wave makes with the direction of the airplane's motion? (b) If the plane is flying at a height of \(6500 \mathrm{~m}\), how long after it is directly overhead will a person on the ground hear the shock wave?

What is the intensity of a sound at the pain level of \(120 \mathrm{~dB}\) ? Compare it to that of a whisper at \(20 \mathrm{~dB}\).

Two sound waves have equal displacement amplitudes, but one has 2.6 times the frequency of the other. (a) Which has the greater pressure amplitude and by what factor is it greater? (b) What is the ratio of their intensities?

A tuning fork is set into vibration above a vertical open tube filled with water (Fig. 16-40). The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is \(0.125 \mathrm{~m}\) and again at \(0.395 \mathrm{~m}\). What is the frequency of the tuning fork?
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