# Simple harmonic motion HELP!

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How would I do this question?

The height of the water on a beach can be approximated as simple harmonic motion with a period of 12hours. If the mean water height is 3.5m, the amplitude of the tide is 1.6m, and `high water' occurs at 7am one day, what would you predict the height of the water to be at 11am?

I'm so confused on what to do?

The height of the water on a beach can be approximated as simple harmonic motion with a period of 12hours. If the mean water height is 3.5m, the amplitude of the tide is 1.6m, and `high water' occurs at 7am one day, what would you predict the height of the water to be at 11am?

I'm so confused on what to do?

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#2

The mean height is the equilibrium position above the ground, so max displacement above the ground is 5.1m and min displacement is 1.9m. It is initially at max displacement (5.1m above the ground). To work around this problem use the equation x=Acos(omega x t) and add 3.5 to your answer

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#3

You model shm using either x = asinwt or x= acoswt depending on the question. Draw a graph where you have time t on the x-axis, and displacement x on the y-axis. Regardless of time, the average water height is 3.5, so as you can imagine, the water will go up and down above and below 3.5. Because of this, you can treat the 3.5 water level as the base level. Then, the amplitude of the tide is 1.6, so this tells you the maximum height the tide can go above 3.5, and the maximum height the tide can go BELOW 3.5. On your y axis you can write that the maximum value of displacement is (3.5+1.6)m, and then the minimum is (3.5-1.6)m.

So imagine that you are being the system, you are looking at a sea wall and the water level is rising up and down non stop to its highest point of 5.1m, and gradually falls to 1.9m (no lower than that) and back and forth and the mid point of this range of heights, by symmetry, is in the centre at 3.5m. Now, do you model the graph as cos or sin? 'High water' occurs at 7 am, so I think that means that you can say that the system 'starts off' when the water is at 5.1m, so you can therefore model it as x = acoswt as, like the graph y = cosx, the maximum value of y occurs at x = 0. Even though it is at 7am, ignore that and just treat this as if it was x=0 on y= cosx.

So the graph is x = 1.6coswt because the 'a' stands for amplitude, which is 1.6. Now w is omega which is found using Period = (2pi)/w. Period is 12 hours, so (2pi)/w = 12. That means w = pi / 6

Overall your graph is therefore x = 1.6cos(pi/6)t

Predicting the height of the water at 11am just means finding x (the displacement) after 4 hours from the start (4 hours after 7) so this is what I meant by ignoring the fact that it's 7am because if we take 7am as t=0 (when the system 'starts off') on the x-axis of the graph, then 11am is t=4.

so at 11am, x = 1.6cos(4pi/6) = -0.8m

Remember displacement takes into account direction, so if you can imagine, if the water level ROSE above 3.5m by 0.8m, that would mean that the water level would now be 4.3m because the displacement moved in the positive direction, but as x = -0.8m that means that the water level lowered (in the negative direction). But if we remember, 3.5m is the average point so the water level is (3.5 - 0.8)m in the instant when t=4 (because as we know displacement is always changing with time). Therefore in the instant when t=4 (at 11am) the water level is 0.8m below the base level, therefore x= 2.7m.

Lots of explaining, I could've done it in 2 mins but I wanted to explain the logic

So imagine that you are being the system, you are looking at a sea wall and the water level is rising up and down non stop to its highest point of 5.1m, and gradually falls to 1.9m (no lower than that) and back and forth and the mid point of this range of heights, by symmetry, is in the centre at 3.5m. Now, do you model the graph as cos or sin? 'High water' occurs at 7 am, so I think that means that you can say that the system 'starts off' when the water is at 5.1m, so you can therefore model it as x = acoswt as, like the graph y = cosx, the maximum value of y occurs at x = 0. Even though it is at 7am, ignore that and just treat this as if it was x=0 on y= cosx.

So the graph is x = 1.6coswt because the 'a' stands for amplitude, which is 1.6. Now w is omega which is found using Period = (2pi)/w. Period is 12 hours, so (2pi)/w = 12. That means w = pi / 6

Overall your graph is therefore x = 1.6cos(pi/6)t

Predicting the height of the water at 11am just means finding x (the displacement) after 4 hours from the start (4 hours after 7) so this is what I meant by ignoring the fact that it's 7am because if we take 7am as t=0 (when the system 'starts off') on the x-axis of the graph, then 11am is t=4.

so at 11am, x = 1.6cos(4pi/6) = -0.8m

Remember displacement takes into account direction, so if you can imagine, if the water level ROSE above 3.5m by 0.8m, that would mean that the water level would now be 4.3m because the displacement moved in the positive direction, but as x = -0.8m that means that the water level lowered (in the negative direction). But if we remember, 3.5m is the average point so the water level is (3.5 - 0.8)m in the instant when t=4 (because as we know displacement is always changing with time). Therefore in the instant when t=4 (at 11am) the water level is 0.8m below the base level, therefore x= 2.7m.

Lots of explaining, I could've done it in 2 mins but I wanted to explain the logic

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(Original post by

You model shm using either x = asinwt or x= acoswt depending on the question. Draw a graph where you have time t on the x-axis, and displacement x on the y-axis. Regardless of time, the average water height is 3.5, so as you can imagine, the water will go up and down above and below 3.5. Because of this, you can treat the 3.5 water level as the base level. Then, the amplitude of the tide is 1.6, so this tells you the maximum height the tide can go above 3.5, and the maximum height the tide can go BELOW 3.5. On your y axis you can write that the maximum value of displacement is (3.5+1.6)m, and then the minimum is (3.5-1.6)m.

So imagine that you are being the system, you are looking at a sea wall and the water level is rising up and down non stop to its highest point of 5.1m, and gradually falls to 1.9m (no lower than that) and back and forth and the mid point of this range of heights, by symmetry, is in the centre at 3.5m. Now, do you model the graph as cos or sin? 'High water' occurs at 7 am, so I think that means that you can say that the system 'starts off' when the water is at 5.1m, so you can therefore model it as x = acoswt as, like the graph y = cosx, the maximum value of y occurs at x = 0. Even though it is at 7am, ignore that and just treat this as if it was x=0 on y= cosx.

So the graph is x = 1.6coswt because the 'a' stands for amplitude, which is 1.6. Now w is omega which is found using Period = (2pi)/w. Period is 12 hours, so (2pi)/w = 12. That means w = pi / 6

Overall your graph is therefore x = 1.6cos(pi/6)t

Predicting the height of the water at 11am just means finding x (the displacement) after 4 hours from the start (4 hours after 7) so this is what I meant by ignoring the fact that it's 7am because if we take 7am as t=0 (when the system 'starts off') on the x-axis of the graph, then 11am is t=4.

so at 11am, x = 1.6cos(4pi/6) = -0.8m

Remember displacement takes into account direction, so if you can imagine, if the water level ROSE above 3.5m by 0.8m, that would mean that the water level would now be 4.3m because the displacement moved in the positive direction, but as x = -0.8m that means that the water level lowered (in the negative direction). But if we remember, 3.5m is the average point so the water level is (3.5 - 0.8)m in the instant when t=4 (because as we know displacement is always changing with time). Therefore in the instant when t=4 (at 11am) the water level is 0.8m below the base level, therefore x= 2.7m.

Lots of explaining, I could've done it in 2 mins but I wanted to explain the logic

**Elliott M**)You model shm using either x = asinwt or x= acoswt depending on the question. Draw a graph where you have time t on the x-axis, and displacement x on the y-axis. Regardless of time, the average water height is 3.5, so as you can imagine, the water will go up and down above and below 3.5. Because of this, you can treat the 3.5 water level as the base level. Then, the amplitude of the tide is 1.6, so this tells you the maximum height the tide can go above 3.5, and the maximum height the tide can go BELOW 3.5. On your y axis you can write that the maximum value of displacement is (3.5+1.6)m, and then the minimum is (3.5-1.6)m.

So imagine that you are being the system, you are looking at a sea wall and the water level is rising up and down non stop to its highest point of 5.1m, and gradually falls to 1.9m (no lower than that) and back and forth and the mid point of this range of heights, by symmetry, is in the centre at 3.5m. Now, do you model the graph as cos or sin? 'High water' occurs at 7 am, so I think that means that you can say that the system 'starts off' when the water is at 5.1m, so you can therefore model it as x = acoswt as, like the graph y = cosx, the maximum value of y occurs at x = 0. Even though it is at 7am, ignore that and just treat this as if it was x=0 on y= cosx.

So the graph is x = 1.6coswt because the 'a' stands for amplitude, which is 1.6. Now w is omega which is found using Period = (2pi)/w. Period is 12 hours, so (2pi)/w = 12. That means w = pi / 6

Overall your graph is therefore x = 1.6cos(pi/6)t

Predicting the height of the water at 11am just means finding x (the displacement) after 4 hours from the start (4 hours after 7) so this is what I meant by ignoring the fact that it's 7am because if we take 7am as t=0 (when the system 'starts off') on the x-axis of the graph, then 11am is t=4.

so at 11am, x = 1.6cos(4pi/6) = -0.8m

Remember displacement takes into account direction, so if you can imagine, if the water level ROSE above 3.5m by 0.8m, that would mean that the water level would now be 4.3m because the displacement moved in the positive direction, but as x = -0.8m that means that the water level lowered (in the negative direction). But if we remember, 3.5m is the average point so the water level is (3.5 - 0.8)m in the instant when t=4 (because as we know displacement is always changing with time). Therefore in the instant when t=4 (at 11am) the water level is 0.8m below the base level, therefore x= 2.7m.

Lots of explaining, I could've done it in 2 mins but I wanted to explain the logic

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#5

(Original post by

Thank you so much, your explanation pretty much cleared up everything for me. So again much thanks

**Yatayyat**)Thank you so much, your explanation pretty much cleared up everything for me. So again much thanks

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(Original post by

No worries, I did A Level further maths in 2016-2018 and did Mechanics M3, are you doing the new A Level further maths syllabus or something?

**Elliott M**)No worries, I did A Level further maths in 2016-2018 and did Mechanics M3, are you doing the new A Level further maths syllabus or something?

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#7

(Original post by

Yes, I’m doing the new spec right now for FM, although I just would’ve hoped the old spec had just carried on for one more year for me. Lol

**Yatayyat**)Yes, I’m doing the new spec right now for FM, although I just would’ve hoped the old spec had just carried on for one more year for me. Lol

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