iheartmaths
New member
- Joined
- Mar 26, 2021
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calculate values of t in an interval for which the value x(t) is zero
The function is x(t)=e-ktcos(2pift)
In this case, f= 3 Hz and k=0.7 m s-1
Putting f in gives cos(6pit)=0
The interval is between t=0 and t=1.
From the graph given, x(t)=0 at six points in this interval. (graph plotting software gives 1/12, 3/12 up to 11/12)
From what I know:
the exponential can only be 0 at infinity so it can be ignored.
The period is 2pi/|b|; or 2pi/6pi in this case, giving 1/3.
I also know cos(x)=0 when x=npi/2, where n is an odd integer.
At this point it has thrown me a bit, just having an issue marrying it all together. I was tempted to arccos both sides which removes cos from the LHS, giving 6pit=1/2pi and then dividing both sides by 6pi to give 1/12 but that only gives one result and I get the feeling that this calculation is a bit misleading.
I've got a feeling I'm missing something obvious an will probably kick myself lol. A push in the right direction will be very much appreciated!
Thanks in advance
The function is x(t)=e-ktcos(2pift)
In this case, f= 3 Hz and k=0.7 m s-1
Putting f in gives cos(6pit)=0
The interval is between t=0 and t=1.
From the graph given, x(t)=0 at six points in this interval. (graph plotting software gives 1/12, 3/12 up to 11/12)
From what I know:
the exponential can only be 0 at infinity so it can be ignored.
The period is 2pi/|b|; or 2pi/6pi in this case, giving 1/3.
I also know cos(x)=0 when x=npi/2, where n is an odd integer.
At this point it has thrown me a bit, just having an issue marrying it all together. I was tempted to arccos both sides which removes cos from the LHS, giving 6pit=1/2pi and then dividing both sides by 6pi to give 1/12 but that only gives one result and I get the feeling that this calculation is a bit misleading.
I've got a feeling I'm missing something obvious an will probably kick myself lol. A push in the right direction will be very much appreciated!
Thanks in advance
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