Find the value of 't' when velocity is equal to 0

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Dinoduck94

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In my coursework, we are given an equation to represent a beam of wood being 'twunged'. The equation represents the displacement of the tip of the beam over time.
We are asked to find the value of 't' when the velocity of the ruler is equal to 0.
I believe I have calculated this however I need some confirmation that my thinking is correct; as I have 't' being equal to infinity seconds when the velocity is 0m/s... it makes sense in my head mathematically but this of course wouldn't be true in the real world.

So the equation is as follows:
x(t) = e^(-kt) * cos(2?ft)
Where 'k' = 0.7, and 'f' = 3 Hz

As this is an expression of displacement over time, I differentiated it with respect to 't' to get velocity over time:
x(t)' = v(t) = -e^(-kt) * (k*cos(2?ft) + 2?f*sin(2?ft))

I understand that the absolute value of the velocity is essentially only the -e^(-kt) part of the expression as this represents the degrading velocity over time.

So by rearranging the -e^(-kt) to make 't' the product, I get the below:

t = ln(0)/(k*ln(-e))

Which then equals:
t = -∞ / (-0.7*ln(-e))

So, because the negatives cancel each other out, therefore t = ∞ when the velocity is 0.
So once the beam is 'twunged' it will never truly stop - although the velocity and the distance (that the tip moves) would exponentially degrade, it would never actually equal 0.


Is this correct, or is my thinking/method wrong?

Thanks
 
I don't see how you "solved" -e^(-kt) * (k*cos(2?ft) + 2?f*sin(2?ft))= 0 and got "ln(0)/kln(-e)" (especially since neither "ln(0)" nor "ln(-e)" exist!). The product of two numbers is 0 if and only if at least one of them is 0. Since e^(-kt) is NEVER 0, we must have k*cos(2?ft) + 2?f*sin(2?ft)= 0 which is the same as sin(2?ft)/cos(2?ft)= tan(2?ft)= -k/2?ft. t= arctan(-k/2?ft)/2?ft.
 
HallsofIvy said:
I don't see how you "solved" -e^(-kt) * (k*cos(2?ft) + 2?f*sin(2?ft))= 0 and got "ln(0)/kln(-e)" (especially since neither "ln(0)" nor "ln(-e)" exist!). The product of two numbers is 0 if and only if at least one of them is 0. Since e^(-kt) is NEVER 0, we must have k*cos(2?ft) + 2?f*sin(2?ft)= 0 which is the same as sin(2?ft)/cos(2?ft)= tan(2?ft)= -k/2?ft. t= arctan(-k/2?ft)/2?ft.
Click to expand...

Hi HallsofIvy, thanks for the response.

The thought process was v = -e^(-kt) and I need to find 't' when 'v' is equal to 0 - so therefore 0 = -e^(-kt).
When I re-arrange this using the formula -kt = x = ln(a) / ln(b), where 'a' would equal '0' in this case and 'b' would equal '-e', I get:

-kt = ln(0) / ln(-e)
Therefore, t = ln(0) / -kln(-e)

I've been taught that the natural logarithm of '0' is undefined, but is commonly represented as '-∞'; and as infinity divided by anything is still infinity, that is where I got my answer.


In the above, why do you state that e^(-kt) is never equal to 0?

The k*cos(2?ft) + 2?f*sin(2?ft) part of the equation is just expressing the frequency of the waveform; it's the -e^(-kt) part that represents the amplitude (which we need to equal 0) - right?
 
HallsofIvy said:
I state it because it is true! Do not know what e^{-kt} is? Do you know what its graph looks like?
Click to expand...

The graph represent a typical exponential decaying curve - with an asymptote along the 'x' axis which tends towards infinity - as per my answer?
 
No, "infinity" is not a number! You cannot say "x= infinity", only that "x goes to infinity" which means it gets larger and larger without bound. e^{-x} (or e^{kx} for any k never crosses the x-axis. That is who I can say that e^(-kt) is never 0.
 
HallsofIvy said:
No, "infinity" is not a number! You cannot say "x= infinity", only that "x goes to infinity" which means it gets larger and larger without bound. e^{-x} (or e^{kx} for any k never crosses the x-axis. That is who I can say that e^(-kt) is never 0.
Click to expand...

Okay, so regardless of the simantics about whether it's "x = infinity" or "x goes to infinity", is my thinking still correct? That once the beam is set into motion, that it doesn't truly come to rest?

Thanks
 
Now that you understand my statement that "e^(-kt) is NEVER 0 what was wrong with my first response?
 
HallsofIvy said:
Now that you understand my statement that "e^(-kt) is NEVER 0 what was wrong with my first response?
Click to expand...

The question asks for the value of 't' when the velocity of the beam is equal to 0 - as the exponential decay of the velocity is represented by e^(-kt), it is this part of the expression that should be used to answer the question.

The k*cos(2?ft) + 2?f*sin(2?ft) part of the expression ultimately defines the shape and frequency of the waveform.
There are numerous values for 't' when 0 = k*cos(2?ft) + 2?f*sin(2?ft) as it will cross the x-axis twice during every cycle of the waveform as it goes from positive to a negative velocity.
 
Last edited:
Dinoduck94 said:
The question asks for the value of 't' when the velocity of the beam is equal to 0 - as the exponential decay of the velocity is represented by e^(-kt), it is this part of the expression that should be used to answer the question.
Click to expand...
No, no, no, no! The question was when the velocity is 0. The fact the absolute value of the velocity is decreasing is irrelevant to this problem. Again, since the exponential is never 0, it is completely irrelevant to this question.

The k*cos(2?ft) + 2?f*sin(2?ft) part of the expression ultimately defines the shape and frequency of the waveform.
There are numerous values for 't' when 0 = k*cos(2?ft) + 2?f*sin(2?ft) as it will cross the x-axis twice during every cycle of the waveform as it goes from positive to a negative velocity.
Click to expand...
Yes, and every time it "goes from positive to negative velocity" it is 0. And, once again, that was the question asked.
 
HallsofIvy said:
No, no, no, no! The question was when the velocity is 0. The fact the absolute value of the velocity is decreasing is irrelevant to this problem. Again, since the exponential is never 0, it is completely irrelevant to this question.


Yes, and every time it "goes from positive to negative velocity" it is 0. And, once again, that was the question asked.
Click to expand...

The coursework asks for a single value of 't', it doesn't ask for a series of values or a formula to represent 't'.
In the past for this module, where multiple values are correct, the question is worded to clarify which value is wanted. For example, which values of 't' between 1 and 2 seconds does v = 0. This clarification is not here in this case.

Thanks for trying to help, but I don't believe what you're stating is what the question is asking. I'll have to go back to my university and ask them to clarify.

Thanks
 
Then what was the question? In your initial post you said "We are asked to find the value of 't' when the velocity of the ruler is equal to 0." You then calculated the velocity to be -e^(-kt) * (k*cos(2?ft) + 2?f*sin(2?ft)). You want that to be 0. For some reason you decided to ignore the "-(k*cos(2?ft) + 2?f*sin(2?ft))" part and try to solve e^(-kt)= 0. That never happens! Since an exponential is never 0, the velocity will be 0 when k*cos(2?ft) + 2?f*sin(2?ft)= 0. That happens repeatedly! Perhaps you are looking for the first time it happens, the smallest value of t that satisfies that equation?
 
HallsofIvy said:
Then what was the question? In your initial post you said "We are asked to find the value of 't' when the velocity of the ruler is equal to 0." You then calculated the velocity to be -e^(-kt) * (k*cos(2?ft) + 2?f*sin(2?ft)). You want that to be 0. For some reason you decided to ignore the "-(k*cos(2?ft) + 2?f*sin(2?ft))" part and try to solve e^(-kt)= 0. That never happens! Since an exponential is never 0, the velocity will be 0 when k*cos(2?ft) + 2?f*sin(2?ft)= 0. That happens repeatedly! Perhaps you are looking for the first time it happens, the smallest value of t that satisfies that equation?
Click to expand...

I have spoken to my university and they have advised that there was wording missing from the question.

The question is now:
'At what value of 't' after the start does the velocity of the tip first become zero?'
Which matches your suggestion above.


I worked through it using the (k*cos(2?ft) + 2?f*sin(2?ft)) part of the equation and my final formula is:

t = (arctan(-k/(2?f))+?n)/2?f
Which matches your post above.

Giving me a final answer of 0.1646974375 seconds; rounded to 0.165 seconds (3 s.f).

Thanks for your help.
 
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