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Sine Wave Tidal written question [closed]
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would appreciate some help working out the sine equation for the following question please:
Depth of water is 6m at low tide and 16m at high tide, which is 6 hours later. Assuming the motion of the water is a simple harmonic, draw a graph to show how the water varies with time over a period of 12 hours, commencing at low tide.
trigonometry
asked Jan 4 at 7:48
DoccurDoccur
31
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closed as off-topic by Aloizio Macedo♦ Jan 5 at 6:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Aloizio Macedo
If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
would appreciate some help working out the sine equation for the following question please:
Depth of water is 6m at low tide and 16m at high tide, which is 6 hours later. Assuming the motion of the water is a simple harmonic, draw a graph to show how the water varies with time over a period of 12 hours, commencing at low tide.
trigonometry
asked Jan 4 at 7:48
DoccurDoccur
31
$endgroup$
closed as off-topic by Aloizio Macedo♦ Jan 5 at 6:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Aloizio Macedo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24
add a comment |
$begingroup$
would appreciate some help working out the sine equation for the following question please:
Depth of water is 6m at low tide and 16m at high tide, which is 6 hours later. Assuming the motion of the water is a simple harmonic, draw a graph to show how the water varies with time over a period of 12 hours, commencing at low tide.
trigonometry
asked Jan 4 at 7:48
DoccurDoccur
31
$endgroup$
would appreciate some help working out the sine equation for the following question please:
Depth of water is 6m at low tide and 16m at high tide, which is 6 hours later. Assuming the motion of the water is a simple harmonic, draw a graph to show how the water varies with time over a period of 12 hours, commencing at low tide.
trigonometry
trigonometry
asked Jan 4 at 7:48
DoccurDoccur
31
asked Jan 4 at 7:48
DoccurDoccur
31
asked Jan 4 at 7:48
DoccurDoccur
31
asked Jan 4 at 7:48
DoccurDoccur
31
asked Jan 4 at 7:48
DoccurDoccur
31
31
closed as off-topic by Aloizio Macedo♦ Jan 5 at 6:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Aloizio Macedo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Aloizio Macedo♦ Jan 5 at 6:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Aloizio Macedo
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24
add a comment |
1 Answer
1
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oldest
votes
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We have the general form
$$y = asin b(t-d)+c$$
where:
$vert avert$ is the amplitude.
$b$ is the frequency.
$(d, c)$ is the horizontal and vertical shift.
Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:
$a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.- The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.
- It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:
$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$
- The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.
Adding all this together, you get
$$y = 5sinfrac{pi}{6}(x-3)+11$$
which becomes
$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$
If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use
$$y = -5cosleft(frac{pi}{6}xright)+11$$
Here are the plots of the graphs.
answered Jan 4 at 8:22
KM101KM101
5,9131523
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have the general form
$$y = asin b(t-d)+c$$
where:
$vert avert$ is the amplitude.
$b$ is the frequency.
$(d, c)$ is the horizontal and vertical shift.
Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:
$a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.- The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.
- It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:
$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$
- The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.
Adding all this together, you get
$$y = 5sinfrac{pi}{6}(x-3)+11$$
which becomes
$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$
If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use
$$y = -5cosleft(frac{pi}{6}xright)+11$$
Here are the plots of the graphs.
answered Jan 4 at 8:22
KM101KM101
5,9131523
$endgroup$
add a comment |
$begingroup$
We have the general form
$$y = asin b(t-d)+c$$
where:
$vert avert$ is the amplitude.
$b$ is the frequency.
$(d, c)$ is the horizontal and vertical shift.
Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:
$a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.- The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.
- It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:
$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$
- The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.
Adding all this together, you get
$$y = 5sinfrac{pi}{6}(x-3)+11$$
which becomes
$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$
If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use
$$y = -5cosleft(frac{pi}{6}xright)+11$$
Here are the plots of the graphs.
answered Jan 4 at 8:22
KM101KM101
5,9131523
$endgroup$
add a comment |
$begingroup$
We have the general form
$$y = asin b(t-d)+c$$
where:
$vert avert$ is the amplitude.
$b$ is the frequency.
$(d, c)$ is the horizontal and vertical shift.
Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:
$a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.- The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.
- It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:
$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$
- The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.
Adding all this together, you get
$$y = 5sinfrac{pi}{6}(x-3)+11$$
which becomes
$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$
If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use
$$y = -5cosleft(frac{pi}{6}xright)+11$$
Here are the plots of the graphs.
answered Jan 4 at 8:22
KM101KM101
5,9131523
$endgroup$
We have the general form
$$y = asin b(t-d)+c$$
where:
$vert avert$ is the amplitude.
$b$ is the frequency.
$(d, c)$ is the horizontal and vertical shift.
Ignoring units for now, at low tide, the depth is $6$ and at high tide, the depth is $16$. Therefore, the middle point/equilibrium point is at $frac{16+6}{2} = frac{22}{2} = 11$. From here, you can deduce that:
$a$ will be the height of the crest or the depth of the trough relative to the equilibrium point. Hence, $a = 16-11 = 5$ or $a = vert 6-11vert = vert -5vert = 5$.- The graph is shifted vertically. Initially, the low-tide would be located at $-5$ (negative amplitude) but it’s now at $+6$ (low-tide), so $c = 11$.
- It takes $6$ hours to go from low-tide to high-tide. This means it would take $6$ hours to go back to low-tide and complete the cycle. Hence, there is one cycle in $12$ hours. We will treat $12$ as our $2pi$ in normal trigonometry, so we can obtain $b$ through a ratio:
$$frac{1}{12} = frac{b}{2pi} iff b = frac{pi}{6}$$
- The graph is shifted horizontally. In a regular sine graph would start from the point of equilibrium. Here, however, you start from the low-tide. The shift must be by $frac{T}{4} = 3$, so $c = 3$.
Adding all this together, you get
$$y = 5sinfrac{pi}{6}(x-3)+11$$
which becomes
$$y = 5sinleft(frac{pi}{6}x-frac{pi}{2}right)+11$$
If you’ve studied cosine graphs as well, you would notice $c = 3$ is really just $frac{T}{4}$ (a quarter of the period), and the shift $left(x-frac{T}{4}right)$ would produce a negative cosine graph, so you could also use
$$y = -5cosleft(frac{pi}{6}xright)+11$$
Here are the plots of the graphs.
answered Jan 4 at 8:22
KM101KM101
5,9131523
edited Jan 4 at 9:09
answered Jan 4 at 8:22
KM101KM101
5,9131523
answered Jan 4 at 8:22
KM101KM101
5,9131523
answered Jan 4 at 8:22
KM101KM101
5,9131523
5,9131523
add a comment |
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 5 at 6:24