Simple harmonic motion rise and fall of water

AI Thread Summary
The discussion revolves around the simple harmonic motion of water in a harbor, with water depths fluctuating between 5.0 m and 9.0 m over a 12-hour cycle. To determine how long a ship requiring a minimum depth of 6.0 m must wait at low tide, it is noted that the amplitude of the tide is 2 meters. The ship can enter the harbor when the tide rises to at least 6.0 m, which occurs when the water level is 1 meter above the lowest position. Clarification was sought on a specific explanation regarding the oscillator's position, but the questioner ultimately understood the concept. The key takeaway is that the ship must wait until the tide rises sufficiently for safe passage.
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Homework Statement



The rise and fall of water in a harbour is simple harmonic. The depth of water varies between 5.0 m at low tide and 9.0 m at high tide. The time between sucessive low tides is 12 hours. A ship, which requires a minimum depth of 6.0 m approaches the harbour at low tide, how long the ship has to wait before entering the harbour?

Homework Equations





The Attempt at a Solution



A hint to start?
 
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The time period of oscillations is given to you as 12hrs, the amplitude is (9-5)/2 = 2 meters, now for the tide to be at least 6m (minimum need for ship to pass), your oscillator must be 1m away from the lowest position moving toward the center.
 
Shivpal said:
The time period of oscillations is given to you as 12hrs, the amplitude is (9-5)/2 = 2 meters, now for the tide to be at least 6m (minimum need for ship to pass), your oscillator must be 1m away from the lowest position moving toward the center.

Thanks Shivpal, but i don get the meaning of this sentence.

'your oscillator must be 1m away from the lowest position moving toward the center.'
 
thereddevils said:
Thanks Shivpal, but i don get the meaning of this sentence.

'your oscillator must be 1m away from the lowest position moving toward the center.'


Never mind, i got it.
 
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