Applications of Trigonometric Functions

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eutas1

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Please refer to the attachments:

I do not really understand this question, and I am stuck on how to do it. What does "in the first cycle of its motion" mean? From 0 to 2pi because it is sine???

When I attempted this question, I understood that you would need to differentiate the displacement function to get the velocity function, where you would then set the velocity function equal to 0 because that is when the particle would be stationary (refer to attachment B). However, looking at the worked solution, I do not understand the simplification of the velocity function that they do (refer to red arrow with question mark in attachment). I also do not understand what they do from then onwards...

Thank you!
 

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[MATH]v(t) = 2\pi \cos(2\pi t) + 2\pi \cos(\pi t) = 0[/MATH]
[MATH]v(t) = 2\pi [\cos(2\pi t) + \cos(\pi t)] = 0 \implies \cos(2\pi t) + \cos(\pi t) = 0[/math]
using the double angle identity for cosine ...

[math]2\cos^2(\pi t) + \cos(\pi t) - 1 = 0[/MATH]
[MATH][2\cos(\pi t) -1] \cdot [\cos(\pi t) + 1] = 0[/MATH]
set each factor = 0 and solve for t

period (one cycle), [MATH]T = \frac{2\pi}{\pi} = 2[/MATH]
 
eutas1 said:
Please refer to the attachments:

I do not really understand this question, and I am stuck on how to do it. What does "in the first cycle of its motion" mean? From 0 to 2pi because it is sine???

When I attempted this question, I understood that you would need to differentiate the displacement function to get the velocity function, where you would then set the velocity function equal to 0 because that is when the particle would be stationary (refer to attachment B). However, looking at the worked solution, I do not understand the simplification of the velocity function that they do (refer to red arrow with question mark in attachment). I also do not understand what they do from then onwards...

Thank you!
Click to expand...
"in the first cycle of its motion": if the function has period P, then the first cycle is the interval from 0 to P.
If the function were sin(x), then the first cycle would be from 0 to 2π. But the function in question is not just sin(x) or cos(x) - it's more complex.
The red arrow part: they divided both sides by 2`pi`.
 
skeeter said:
[MATH]v(t) = 2\pi \cos(2\pi t) + 2\pi \cos(\pi t) = 0[/MATH]
[MATH]v(t) = 2\pi [\cos(2\pi t) + \cos(\pi t)] = 0 \implies \cos(2\pi t) + \cos(\pi t) = 0[/math]
Click to expand...

Oh yeah! I understand that part now.

skeeter said:
using the double angle identity for cosine ...

[math]2\cos^2(\pi t) + \cos(\pi t) - 1 = 0[/MATH]
[MATH][2\cos(\pi t) -1] \cdot [\cos(\pi t) + 1] = 0[/MATH]
set each factor = 0 and solve for t

period (one cycle), [MATH]T = \frac{2\pi}{\pi} = 2[/MATH]
Click to expand...

We have not learnt the double angle identity for cosine... Is there another way to do the question that does not involve this method?
 
lev888 said:
"in the first cycle of its motion": if the function has period P, then the first cycle is the interval from 0 to P.
If the function were sin(x), then the first cycle would be from 0 to 2π. But the function in question is not just sin(x) or cos(x) - it's more complex.
The red arrow part: they divided both sides by 2`pi`.
Click to expand...

Ah, I see I see.
Thank you!
 
eutas1 said:
Oh yeah! I understand that part now.



We have not learnt the double angle identity for cosine... Is there another way to do the question that does not involve this method?
Click to expand...

[MATH]\cos(2\pi t) = \cos(\pi t + \pi t) = \cos(\pi t)\cos(\pi t) - \sin(\pi t)\sin(\pi t) = \cos^2(\pi t) - \sin^2(\pi t) = \cos^2(\pi t) - [1 - \cos^2(\pi t)] = 2\cos^2(\pi t) - 1[/MATH]
now you know the double angle identity for cosine
 
skeeter said:
[MATH]\cos(2\pi t) = \cos(\pi t + \pi t) = \cos(\pi t)\cos(\pi t) - \sin(\pi t)\sin(\pi t) = \cos^2(\pi t) - \sin^2(\pi t) = \cos^2(\pi t) - [1 - \cos^2(\pi t)] = 2\cos^2(\pi t) - 1[/MATH]
now you know the double angle identity for cosine
Click to expand...

How is cos(πt+πt) = cos(πt)cos(πt) − sin(πt)sin(πt) ???
 
eutas1 said:
Oh, it is because we have not learnt sum and difference identities either...
Click to expand...

In that case, you'll have to use the "trial and error" method of solving a trig equation as shown in your initial image ... good luck with that.
 
eutas1 said:
Please refer to the attachments:

I do not really understand this question, and I am stuck on how to do it. What does "in the first cycle of its motion" mean? From 0 to 2pi because it is sine???

When I attempted this question, I understood that you would need to differentiate the displacement function to get the velocity function, where you would then set the velocity function equal to 0 because that is when the particle would be stationary (refer to attachment B). However, looking at the worked solution, I do not understand the simplification of the velocity function that they do (refer to red arrow with question mark in attachment). I also do not understand what they do from then onwards...

Thank you!
Click to expand...
If you add the same expression to itself twice (or any number of times) and get 0, then the expression must be 0. That is, if x+x=0, then x=0.
 
Jomo said:
If you add the same expression to itself twice (or any number of times) and get 0, then the expression must be 0. That is, if x+x=0, then x=0.
Click to expand...

The same expression is not added to itself twice though...

1618826022641.png
 
eutas1 said:
But cos(2pi*t) is not equal to cos(pi*t)?
Click to expand...
This is true, but why mention it?

If you have 7xyz + 7xw = 0, you CAN divide all terms by 7x and get yz + w = 0. Note that yz does not equal w. In fact, yz=-w.
 
Jomo said:
This is true, but why mention it?

If you have 7xyz + 7xw = 0, you CAN divide all terms by 7x and get yz + w = 0. Note that yz does not equal w. In fact, yz=-w.
Click to expand...

But where is the same expression added to itself twice? (post #12)
 
eutas1 said:
But where is the same expression added to itself twice? (post #12)
Click to expand...
You do not have the same expression added to itself twice. You do how common factors. Both terms have 2pi in common. You can divide both sides of the equation (top equation) by 2pi and then get the bottom equation.
 
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