# Simple harmonic motion help pls

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#1
SO I have stuck by the following question, can anyone help to explain?
Its question b thanks
Last edited by sherryeww; 1 year ago
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1 year ago
#2
(Original post by sherryeww)
SO I have stuck by the following question, can anyone help to explain?
Its question b thanks
I cannot seem to see any question.
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#3
here is the question.
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#4
(Original post by Eimmanuel)
I cannot seem to see any question.
I’ve inserted the pic below, thanks!
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1 year ago
#5
(Original post by sherryeww)
I’ve inserted the pic below, thanks!
How far have you got?
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1 year ago
#6
(Original post by sherryeww)
I’ve inserted the pic below, thanks!
Frequency of oscillation = 2pi/4.5 =4pi/9
Amplitude of oscillation = 1.15 m
x = 1.15 cos 4pi t /9
0.6 m means the amplitude is -0.55m

0.55 = 1.15 cos 1.4 t
t = 0.766 s

so time = 4.5 - 2 x 0.766 = 3.0 s
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1 year ago
#7
(Original post by sherryeww)
here is the question.
Try to model the SHM motion using a trigonometric function such as -Acos(ωt), where the barrier starts from the bottom. The height of the barrier moves is 2A.

Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t1 and then at time t2. The time difference between t1 and t2 will be the maximum time.
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1 year ago
#8
(Original post by golgiapparatus31)
Frequency of oscillation = 2pi/4.5 =4pi/9
Amplitude of oscillation = 1.15 m
x = 1.15 cos 4pi t /9
0.6 m means the amplitude is -0.55m

0.55 = 1.15 cos 1.4 t
t = 0.766 s

so time = 4.5 - 2 x 0.766 = 3.0 s

I got a notification that you quoted my post. It seems to say something like the maximum time is the time above 0.6 m. What is wrong with my proposed working?

(Original post by golgiapparatus31)
…0.6 m means the amplitude is -0.55m ....
Are you really sure amplitude can be negative in describing SHM for A-level physics?

https://en.wikiversity.org/wiki/Amplitude

https://en.wikibooks.org/wiki/A-leve...armonic_Motion
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#9
(Original post by golgiapparatus31)
Frequency of oscillation = 2pi/4.5 =4pi/9
Amplitude of oscillation = 1.15 m
x = 1.15 cos 4pi t /9
0.6 m means the amplitude is -0.55m

0.55 = 1.15 cos 1.4 t
t = 0.766 s

so time = 4.5 - 2 x 0.766 = 3.0 s
I’m wondering how do I get 0.55 and 1.15, also why 4.5-2times, can you explain a bit more pls? Many thanks!
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#10
(Original post by Eimmanuel)
Try to model the SHM motion using a trigonometric function such as -Acos(ωt), where the barrier starts from the bottom. The height of the barrier moves is 2A.

Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t1 and then at time t2. The time difference between t1 and t2 will be the maximum time.
I don’t really understand ‘
Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t1 and then at time t2. The time difference between t1 and t2 will be the maximum time’ can you explain a bit more?many thanks!!!
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1 year ago
#11
(Original post by sherryeww)
I don’t really understand ‘
Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t1 and then at time t2. The time difference between t1 and t2 will be the maximum time’ can you explain a bit more?many thanks!!!

If you look at the picture, the black line represents the ground. When I say “Assume that the barrier starts at the bottom or -A,” meaning the bottom of the barrier starts its motion from the ground. (I know the description is bad in the previous post. )

The pink line in the picture shows the position of (-A + 0.60 m) from the ground while the red curve is modelling the motion of the bottom of the barrier. When the bottom of the barrier reaches the position (-A + 0.60 m) for the first time is around 0.7 s and the second time is around 3.7 s as shown by the intersection between the pink line and red curve. The difference between the 2 timings would be the maximum time.
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#12
(Original post by Eimmanuel)

If you look at the picture, the black line represents the ground. When I say “Assume that the barrier starts at the bottom or -A,” meaning the bottom of the barrier starts its motion from the ground. (I know the description is bad in the previous post. )

The pink line in the picture shows the position of (-A + 0.60 m) from the ground while the red curve is modelling the motion of the bottom of the barrier. When the bottom of the barrier reaches the position (-A + 0.60 m) for the first time is around 0.7 s and the second time is around 3.7 s as shown by the intersection between the pink line and red curve. The difference between the 2 timings would be the maximum time.
(Original post by Eimmanuel)

If you look at the picture, the black line represents the ground. When I say “Assume that the barrier starts at the bottom or -A,” meaning the bottom of the barrier starts its motion from the ground. (I know the description is bad in the previous post. )

The pink line in the picture shows the position of (-A + 0.60 m) from the ground while the red curve is modelling the motion of the bottom of the barrier. When the bottom of the barrier reaches the position (-A + 0.60 m) for the first time is around 0.7 s and the second time is around 3.7 s as shown by the intersection between the pink line and red curve. The difference between the 2 timings would be the maximum time.
But I don’t understand why it would reach the position(-A+0.6)?
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1 year ago
#13
(Original post by sherryeww)
But I don’t understand why it would reach the position(-A+0.6)?
I am not sure how can explain this. The question states that "A contestant requires a space at least 0.6 m high to get under the barrier." So the barrier need to moves up by 0.6 m.

If you are at basement level 5, which i called -5, when you move up 3 levels, isn't -5 + 3.
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#14
(Original post by Eimmanuel)
I am not sure how can explain this. The question states that "A contestant requires a space at least 0.6 m high to get under the barrier." So the barrier need to moves up by 0.6 m.

If you are at basement level 5, which i called -5, when you move up 3 levels, isn't -5 + 3.
Thanks I finally get it!!!
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