the answers would have to be given in 2sf right? can’t leave it as fraction?
(i hate decimals -pls help)
Yes, you are right. I think you get another mark for putting the answer in decimals — it says so on the front of the paper. So I guess you should do so. But of course everyone hates arithmetic :-).
Why did you use dy/dt as wA instead of -wA?
Hi Jemisha, thanks for your question. The value of cos(kx-wt) varies between +1 and -1. So dy/dt varies between -wA and +wA respectively. The question asks for the “maximum vertical velocity”, which is wA.
Actully you can also work out the exact value of w and A,or even K,but K is not helpful in get the maximum velocity…
Yes you are absolutely right; thanks for pointing it out. Somehow I managed to fail to spot the sentence in the question that had all the numbers in it. I have added the bottom three lines to the answer, containing the calculation of the actual value.
hi I don’t quite understand your working out, can you explain it to me?
Sure, no problem.
Firstly, we’ve been given an equation for the height y and we are being asked a question about the vertical velocity. The vertical velocity is the differential of the height with respect to time, so it makes sense to differentiate the expression and see what we get.
When we do this, we can treat the expression kx as a constant. If you don’t know this differential immediately, you can work it out by using the trigonometric identity sin(A-B) = sinA cosB – sinB cosA, and differentiating that expression. You will find that the differential then simplifies, using another trig identity, to -wAcos(kx-wt).
Then we can observe that we have been asked for the maximum value of the velocity. We know that all cosines vary between -1 and +1, so we can see that the highest possible value is when cos(kx-wt) is -1, so the highest value is wA.
Is that OK?
ah, that makes sense! Thank you very much!!
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