# Shm help (m3)Watch

#1
I don’t understand the markscheme for part d of this question at all. From my understanding of SHM, you use x = acos(wt) when motion begins at the max displacement. Why have they used x = asin(wt) and a positive distance twice? Any help is really appreciated Question:https://imgur.com/a/42qaK8WMarkschem....com/a/Pkvw1Rp
0
2 months ago
#2
(Original post by supalape)I don’t understand the markscheme for part d of this question at all. From my understanding of SHM, you use x = acos(wt) when motion begins at the max displacement. Why have they used x = asin(wt) and a positive distance twice? Any help is really appreciated Question:https[/s]//imgur.com/a/Pkvw1Rp]https://imgur.com/a/42qaK8WMarkscheme//imgur.com/a/Pkvw1Rp

You're right,
Acos(wt)
is the normal solution. However, for DB really want to calculate the time taken not from the starting point, but from half the amplitude to the centre (not the end to half the amplitude). You could write this in a couple of ways
1) Calculate the quarter period (time from C to B) then subtract the time for CD. The latter part would involve the normal cos, i.e.
l/4 = l/2*cos(w*T_CD)
2) Calculate the time from D to B (same as the time from B to D). This has a shift of 90 (pi/2 rad) and you use the fact that
cos(90-x) = sin(x)
Both views are equivalent. Another way of imagining it is that you have x=Asin(wt) when the initial displacement from the equilbrium is zero, but the initial velocity is non-zero. This is equivalent to what we want as we want a time about the equilibrium point.
1
#3
(Original post by mqb2766)
(Original post by supalape)I don’t understand the markscheme for part d of this question at all. From my understanding of SHM, you use x = acos(wt) when motion begins at the max displacement. Why have they used x = asin(wt) and a positive distance twice? Any help is really appreciated Question:https[/s]//imgur.com/a/Pkvw1Rp]https://imgur.com/a/42qaK8WMarkscheme//imgur.com/a/Pkvw1Rp

You're right,
Acos(wt)
is the normal solution. However, for DB really want to calculate the time taken not from the starting point, but from half the amplitude to the centre (not the end to half the amplitude). You could write this in a couple of ways
1) Calculate the quarter period (time from C to B) then subtract the time for CD. The latter part would involve the normal cos, i.e.
l/4 = l/2*cos(w*T_CD)
2) Calculate the time from D to B (same as the time from B to D). This has a shift of 90 (pi/2 rad) and you use the fact that
cos(90-x) = sin(x)
Both views are equivalent. Another way of imagining it is that you have x=Asin(wt) when the initial displacement from the equilbrium is zero, but the initial velocity is non-zero. This is equivalent to what we want as we want a time about the equilibrium point.
Thank you that makes sense! I still don’t understand why they used a positive distance from B to slack string though, surely it’s negative as they used C to B as positive
0
2 months ago
#4
There are a lot of symmetries in SHM, as it's based on trig curves. The time would be the same whether X is POS or neg.
(Original post by supalape)
Thank you that makes sense! I still don’t understand why they used a positive distance from B to slack string though, surely it’s negative as they used C to B as positive
0
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