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butterflydreamer

  • one year ago

check my answer please :) SIMPLE HARMONIC MOTION - Question: The rise and fall of the ride at the mouth of a river is assumed to be simple harmonic motion. The depth of water at low tide is 0.7m and the depth at high tide is 3.7m. If low tide occurred at 8:55am and high tide at 3:05pm, find the earliest time at which the boat could enter the mouth of the river if it required the water to be at least 2m deep. My solution: http://prntscr.com/7d1ex1

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  1. butterflydreamer
    • one year ago
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    Low tide at 8:55am is 0.7m High tide at 3:05 pm is 3.7m Therefore, centre at 12:00pm will be 2.2m Amplitude = 1.5 Period = 2pi/n = ?

  2. IrishBoy123
    • one year ago
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    i suggest you calculate the period as you have the amplitude the period, T, is time for a complete oscillation. you go from trough to peak during times 8:55am to 3:05 pm. isn't that half an oscillation? once you have T, you can then use \( \omega = \frac{2 \pi}{T}\) in your shm equation: \(y - y_0 = A cos( \omega t + \phi)\), if that is how you are presenting it.

  3. butterflydreamer
    • one year ago
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    ohhh okayyy! Give me a minute or so to work it out :)

  4. butterflydreamer
    • one year ago
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    wait.. if 8:55am - 3:05 pm is HALF an oscillation.. that is 6hrs and 10 minutes... So it takes 12 hrs and 20 minutes for a complete oscillation.. Therefore w = 740 (12hrs10 mins) Sooo... 2pi/n = 740 n = pi/370 |dw:1433425233117:dw| |dw:1433425384627:dw|

  5. IrishBoy123
    • one year ago
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    |dw:1433425997658:dw|

  6. butterflydreamer
    • one year ago
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    oops i wrote that wrong LOL. Every part which says 360 should be 370 :)

  7. IrishBoy123
    • one year ago
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    |dw:1433426031745:dw| yes, but method looks good haven't checked your calcs or answer

  8. IrishBoy123
    • one year ago
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    |dw:1433426180829:dw|

  9. butterflydreamer
    • one year ago
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    soz? LOL Oh dear.. i made so many errors -_- OKAY i will write it out on paper because that's the easiest method.

  10. IrishBoy123
    • one year ago
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    final: i assume adding \(\pi \) on RHS somehow negates the fact it is a negative cosine. only saying because i would need to work it through to get it, and don't have time -- but I suspect it is a little trick that you use regularly....

  11. butterflydreamer
    • one year ago
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    yeah, it is a trick that i found online LOL. Okay basically if i write my working out properly, it will look like this: http://prntscr.com/7d1ex1

  12. IrishBoy123
    • one year ago
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    looks good

  13. butterflydreamer
    • one year ago
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    alrighties thank you so muccch! :)

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