Hello, Below I have attached an image containing the graph and the values given I must find Vavg through Integration, I believed the answer is 5Vm/T can someone verify this?
You know the answer is wrong because the units don't work out. The average value of V has to have units of V, not V/T. Show your work (or describe your approach clearly) and we can probably help you find where you went wrong.
New answer is Vm/3Tsquared first equation is 2Vmt/T second equation is -2Vm/T + 2Vm Vavg = 1/b-a * ∫2Vm/T*t dt|T/2-0 + ∫-2Vm/T*t dt|T-T/2 = 1/T * [(2Vm/T*t^2/2)]| T/2-0 + [(-2Vmt/T *t^2/2 + 2Vmt)]| T-T/2 = 1/T * [(Vm/T*(T/2)^2)] + [(-Vm/T*(T)^2 + 2Vm(T))-(-Vm/T*(T/2)^2 + 2Vm(T/2)) =1/T *[Vm/T * T/4] + [(-Vm/T*T^2) + (2Vm*T))-((-Vm/T *T/4) + (2Vm*T/2))] =1/T *[(VmT/4T)] + [(-VmT + 2VmT)-((-VmT/4T)+(2VmT/2))] =1/T *[Vm/3T] + [(VmT)-(VmT/8T)] =1/T * Vm/3T Vavg = Vm/3T^2
Didn't you read what WBahn has written? The units of Vavg must be V. For example, if we are discussing distance traveled, distance/time is velocity. Average distance traveled must be a distance.
Units still don't work out. You therefore KNOW it is wrong! You are also probably not writing what you mean. As written you are saying: Vavg = (Vm/3)*(T^2) You should also be able to determine the answer without performing the integration, since you know that the integral gives you the area under the curve. Well, what is the area under a triangle that has a base of T and a hight of Vm? The notion of Vavg is simply asking how high does a rectangle that has a width of T need to be to have the same area? I'll agree with the first one, but the units don't work out on the second one, so you KNOW it is wrong. Have no idea what 'b' and 'a' are, or what (1/b) - (a*integral#1) + (integral#2) is supposed to yield. The problem is that you are being very sloppy with your math. You are dropping terms and factors at several points in your computation. At each of those points, a simple units check would catch it and let you correct it before moving on. The vast majority of mistakes you make will screw up the units, but if you don't carry them or don't check them, you forfeit what is perhaps the single best error-detection tool available to an engineer or technician. That's how space probes end up smacking into planets instead of going into orbit around them.
Ok so I've finaly solved the problem and Vavg = Vm/2 which I know its correct (teacher looked at it). So now I'm having issues solving Vrms, the answer I got is square root 10/3*Vm. Can anybody verify that this answer is correct? Thanks
Will you PLEASE use parentheses to show what you are trying to say with your expression? Do you expect Vrms it be more or less than Vm? Do you expect Vrms it be more or less than Vavg?
And what about the parentheses? I can't tell what you intend to be within the squareroot and what is meant to be outside it. I can't tell what you mean to be in the denominator and what is just the next factor.
You have a units problem right on the very first line. The left hand side is (volts) but the right hand side is (1/sec)(volts^2)(sec) which will yield (volts^2). Thus you know you initial set up is wrong. Why go any further? You then get sloppy with the math just a couple lines down. You square an expression that's in parentheses but only square some of the factors, not all of them. Checking your units would have caught this. Most mistakes you make (and you will make them throughout your career, have no illusions about that) will mess up the units. Tracking and checking your units is a simple and effective way to catch such mistakes. As a practicing engineer, failing to use such a simple and effective error detection strategy as part of your normal practices borders on criminal negligence, as far as I'm concerned. Remember, incompetent doctors generally kill people one at a time, incompetent engineers do it in job lots.
You might want to check your limits of integration. Setting up the problem is the most critical phase of the solution. Take the time and effort to be sure it is correct, because if it isn't, then you are just wasting your time from that point on.
You are getting very close. Now, notice that you previously indicated that Vrms should be less than Vm but greater than Vavg. If Vavg is Vm/2, is your current result for Vrms greater than Vavg? Tell you what, you are close enough that if you post the steps you took to get the answer above, I will explain which lines have what problems. Provided you do two things: First, be explicit with your parentheses (and brackets). Remember, saying (1/T) x + y is the same as [(1/T) x] + y and not (1/T) (x + y) If you mean the latter, you must be explicit. Second, on each line, indicate the units associated with each term on both sides of the equal sign. My guess is that if you do these two things, you will discover your remaining error yourself. If not, I will be happy to point out where you are getting tripped up.
Nope. Check units! Also, please, please, please get in the habit of using parentheses! Many people will write 1/(3Vm) as 1/3Vm because they think of the 3 as being tightly bound to the Vm. Many people will interpret it this way when they see it for the same reason. Even some who don't will suspect that is what you meant, otherwise why didn't you write it as Vm/3 to begin with? By writing (1/3)Vm, you make it clear and unambiguous what you meant. Isn't that the goal when communicating?