# Simple harmonic motion help pls

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

SO I have stuck by the following question, can anyone help to explain?

Its question b thanks

Its question b thanks

Last edited by sherryeww; 1 year ago

0

reply

Report

#2

(Original post by

SO I have stuck by the following question, can anyone help to explain?

Its question b thanks

**sherryeww**)SO I have stuck by the following question, can anyone help to explain?

Its question b thanks

0

reply

(Original post by

I cannot seem to see any question.

**Eimmanuel**)I cannot seem to see any question.

0

reply

Report

#5

(Original post by

I’ve inserted the pic below, thanks!

**sherryeww**)I’ve inserted the pic below, thanks!

0

reply

Report

#6

(Original post by

I’ve inserted the pic below, thanks!

**sherryeww**)I’ve inserted the pic below, thanks!

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

0

reply

Report

#7

(Original post by

here is the question.

**sherryeww**)here is the question.

Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t

_{1}and then at time t

_{2}. The time difference between t

_{1}and t

_{2}will be the maximum time.

0

reply

Report

#8

(Original post by

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

**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

…0.6 m means the amplitude is -0.55m ....

**golgiapparatus31**)…0.6 m means the amplitude is -0.55m ....

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

https://en.wikibooks.org/wiki/A-leve...armonic_Motion

0

reply

**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

0

reply

(Original post by

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 t

**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 t

_{1}and then at time t_{2}. The time difference between t_{1}and t_{2}will be the maximum time.Assume that the barrier starts at the bottom or -A, the barrier first reaches (-A + 0.60 m) at time t

_{1}and then at time t

_{2}. The time difference between t

_{1}and t

_{2}will be the maximum time’ can you explain a bit more?many thanks!!!

0

reply

Report

#11

(Original post by

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 t

**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 t

_{1}and then at time t_{2}. The time difference between t_{1}and t_{2}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.

0

reply

(Original post by

If you look at the picture, the black line represents the ground. When I say “Assume that the barrier starts at the bottom or -

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.

**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.

**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.

0

reply

Report

#13

(Original post by

But I don’t understand why it would reach the position(-A+0.6)?

**sherryeww**)But I don’t understand why it would reach the position(-A+0.6)?

If you are at basement level 5, which i called -5, when you move up 3 levels, isn't -5 + 3.

0

reply

(Original post by

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.

**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.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top