   # The Bay of Fundy, Nova Scotia, has the highest tides in the world. Assume in midocean and at the mouth of the bay the Moon’s gravity gradient and the Earth’s rotation make the water surface oscillate with an amplitude of a few centimeters and a period of 12 h 24 min. At the head of the bay, the amplitude is several meters. Assume the bay has a length of 210 km and a uniform depth of 36.1 m. The speed of long-wavelength water waves is given by v = g d , where d is the water’s depth. Argue for or against the proposition that the tide is magnified by standing-wave resonance. ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553292

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

10th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781337553292
Chapter 17, Problem 19P
Textbook Problem
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## The Bay of Fundy, Nova Scotia, has the highest tides in the world. Assume in midocean and at the mouth of the bay the Moon’s gravity gradient and the Earth’s rotation make the water surface oscillate with an amplitude of a few centimeters and a period of 12 h 24 min. At the head of the bay, the amplitude is several meters. Assume the bay has a length of 210 km and a uniform depth of 36.1 m. The speed of long-wavelength water waves is given by v = g d , where d is the water’s depth. Argue for or against the proposition that the tide is magnified by standing-wave resonance.

To determine

The proposition that the tide is magnified by standing wave resonance.

### Explanation of Solution

Given Info: The length of the bay is 210km , the depth of the bay is 36.1m and the speed of the long wavelength water wave is gd .

The speed of the long wavelength water wave is,

v=gd

Here,

v is the speed of the long wavelength water wave.

g is the acceleration due to gravity.

d is the depth of the bay.

Substitute 36.1m for d and 9.81m/s2 to find the v .

v=9.81m/s2×36.1m=18.81m/s

The bay has one end open and one closed. Its simplest resonance is with a node of horizontal velocity, which is also an antinode of vertical displacement at the head of the bay and an antinode of velocity, which is a node of displacement at the mouth of the bay. and the length of the bay represents the displacement at the mouth of the bay. Hence, the displacement as a form of wave at the mouth of the will be equal to the one fourth of the wavelength.

db=λ4 (1)

Here,

λ is the wavelength.

db . is the displacement as a form of wave at the mouth

Substitute 210km for db in equation (1) to find the λ

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