Year 12 Hard Simple Harmonic Motion Question

[katsicum]

Hi, I am getting stuck on the following question. I understand why they advise us to use the cos form instead of the sin form since the particle start at x=1 which is not the origin. However, usually, I would assume that a=1 since it would be the case that the particle starts from rest so at t=0 x=n would be the amplitude that for this question I am not sure. Could someone please guide me through the entire question, not just a couple of hints since I don't think that would be enough. I have literally been sitting here for almost 1 hour straight but i still don't get the proof.

Thank you so much for your help!

 

[Otis]

… I understand why they … use [cosine] instead of [sine] since the particle start at x=1
… I would assume that a=1 since … the particle starts from rest …

Hi katsicum. I'm not sure what "starts at rest" means, in this exercise. Your statements above lead me to wonder if you're thinking in terms of the graph of y=cos(x). In this exercise, the particle is oscillating on the x-axis, as a function of time.

Were we to look at the function's graph, the x-axis would be vertical and the t-axis would be horizontal.

We know the amplitude must be greater than 1 because the particle oscillates more than 5 units away from the Origin. We can also use the given form, to see that a cannot be 1. Set x=1 and t=0.

1 = a · cos(n·0 + α)
1 = a · cos(α)
a = 1/cos(α)

So, α would need to be 0, for a to be 1, but if α were 0 there would be no phase shift, and that leads to a contradiction because x is not at its maximum distance from the Origin when t is 0.

Using the given value of the period (6.7758, rounded), I'm thinking the graph looks something like this (because symmetry indicates a maximum must occur halfway between t=1 and t=2):



No immediate ideas occurred to me about why the period may be expressed in terms of arccos(3/5), but I didn't experiment with the algebra part much. I posted this information because I wanted you to have a better sense of what the function represents, as well as showing a graph.

PS: To form that graph, my value of a is negative. The amplitude is |a|.

?

 

[Deleted member 4993]

Hi, I am getting stuck on the following question. I understand why they advise us to use the cos form instead of the sin form since the particle start at x=1 which is not the origin. However, usually, I would assume that a=1 since it would be the case that the particle starts from rest so at t=0 x=n would be the amplitude that for this question I am not sure. Could someone please guide me through the entire question, not just a couple of hints since I don't think that would be enough. I have literally been sitting here for almost 1 hour straight but i still don't get the proof.

Thank you so much for your help!

View attachment 19699

You had posted a question on the same topic (SHM) at:

https://www.freemathhelp.com/forum/...cs-simple-harmonic-motion.123182/#post-498749

and informed us that you had solved that problem. Please review that problem again and make sure that you had understood the solution (you did NOT share the work with us).

Hint for this problem:

If the equation of the motion is:

y = A * cos (w*t + p) then

T = 2*pi/w and

w = [cos^-1(y/A) - p]/t

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[Otis]

… PS: To form that graph, my value of a is negative. The amplitude is |a|.

Today, I've used a sum identity for cosine, and I found a second function that fits the given information. The value of a is positive, in the second function.

I need to work through Subhotosh's post, however. I may have misinterpreted something because my parameter results don't jive with all of his equations.

Again, I've not generated a proof; I've just confirmed two functions that fit the information (I hope), using the given period as the groundwork.

?

 

[katsicum]

Today, I've used a sum identity for cosine, and I found a second function that fits the given information. The value of a is positive, in the second function.

I need to work through Subhotosh's post, however. I may have misinterpreted something because my parameter results don't jive with all of his equations.

Again, I've not generated a proof; I've just confirmed two functions that fit the information (I hope), using the given period as the groundwork.

?

Hi, could you please give me a hint on how to use the sum identity for cosine for this question? I have tried that but it didn't work for me!

 

[Otis]

… please give me a hint on how to use the sum identity for cosine for this question? …

Hi katsicum. I didn't use that identity to prove the given period. I applied it, to see what would happen, because I recognized ahead of time that at least one cos(α) factor would cancel. After substituting the given arccosine component, the result was an equation to solve for α.

… it didn't work for me!

Were you also trying to find parameter values? What didn't work, specifically?

If you're also trying to solve for α, then here's the equation I started with. My first step was to apply the sum identity for cosine.

\(5 = \frac{1}{\cos(α)} \cdot \cos\left( \arccos\left(\frac{3}{5}\right) + α \right)\)

I won't have time to review Subhotosh's post until tomorrow, and I really need to (before posting again). His formula for n doesn't work with my values, so I probably goofed something.

?

 

[Deleted member 4993]

Hi katsicum. I didn't use that identity to prove the given period. I applied it, to see what would happen, because I recognized ahead of time that at least one cos(α) factor would cancel. After substituting the given arccosine component, the result was an equation to solve for α.


Were you also trying to find parameter values? What didn't work, specifically?

If you're also trying to solve for α, then here's the equation I started with. My first step was to apply the sum identity for cosine.

\(5 = \frac{1}{\cos(α)} \cdot \cos\left( \arccos\left(\frac{3}{5}\right) + α \right)\)

I won't have time to review Subhotosh's post until tomorrow, and I need to before posting again. (His formula for n doesn't work with my values, so I probably goofed something.)?

You say:

"(His formula for n doesn't work......"

I did not have any 'n' (response #3) in any formula there......

 

[Otis]

… I did not have any 'n' (response #3) in any formula there......

That's true. Except for 't', you've changed the given symbols.

Your 'w' represents 'n', correct?

?

 

[Deleted member 4993]

That's true. Except for 't', you've changed the given symbols.

Your 'w' represents 'n', correct?

?

Yes and No - in SHM the term 'w' (omega) represents angular speed.

The particle travels (angular) distance = 2 * pi during one cycle (Period = T)

then we have:

w = 2 * pi/T

I hate to use wiki as reference - but that is the quickest one.

https://en.wikipedia.org/wiki/Simple_harmonic_motion

There you will find: the equation for SHM to be:

That is a standard expression in dynamics.

 

[Otis]

… 'w' (omega) represents angular speed …

Okay, but I was thinking the exercise could be answered by considering math of sinusoidal functions alone, instead of starting with physics (i.e., deriving the standard expression using Hooke's Law to write a differential equation).

Here are the parameter values from one of the two sets that I've obtained.

a ≈ 5.5902
n ≈ 0..9273
α ≈ -1.3909

We're given the (t,x) point (1,5). You posted a formula:

n = [arccos(x/a) - α]/t

0..9273 = [arccos(5/5.5902) + 1.3909]/1
0.9273 = 1.8546

I get the same result, using the other parameter set. In neither case may I use the given point (0,1), in the formula for n.

My functions match the given information, but don't jive with the formula above. I'm not sure what''s wrong, and I still don't understand how to demonstrate the arccos(3/5) term in the given period.

 

[Deleted member 4993]

Hi, I am getting stuck on the following question. I understand why they advise us to use the cos form instead of the sin form since the particle start at x=1 which is not the origin. However, usually, I would assume that a=1 since it would be the case that the particle starts from rest so at t=0 x=n would be the amplitude that for this question I am not sure. Could someone please guide me through the entire question, not just a couple of hints since I don't think that would be enough. I have literally been sitting here for almost 1 hour straight but i still don't get the proof.

Thank you so much for your help!

View attachment 19699

Is the problem "translated" from another language?



I do not understand the statement:

"At the end of each of the first two seconds..."​

I cannot visualize the adverb "each" here - along with "first".