S.H.M and calculating depth of water

[MorallyObtuse]

Greetings,

Kindly, may I have my answers checked, please.

Homework Statement

d. A 40kg child sits on a swing of length 8.0m and is swinging with a maximum displacement of 2.0m from the equilibrium position. If the motion can be considered to be SHM, calculate the:

i. Period of the swing.
ii. Maximum vertical distance the child rises above the equilibrium position
iii. Maximum velocity of the child during the motion
iv. The total energy produced due to this motion

Homework Equations

i. T = 2π√l/g
ii. x = rsinwt
iii. V = coswt
iv. (Kinetic Energy) Ek = 1/2mv²

The Attempt at a Solution

i. From T = 2π√l/g = 2πw, we see that w = √g/l = √9.8/1 = 9.8, assuming g = 9.8 m/s ²
T = √ 2π/9.8
= 0.8 s

ii. x = rsinwt
w = 2πf
f = 1/0.8
= 0.25 Hz
w = 2π x 1.25
= 7.85 1/rads
x = 2m x sin7.85 1/rads x 0.8
= 1.6

iii. V = wrcoswt
= 0.8 1/rads x 2π x cos7.851/rads x 0.8
= 0.02 m/s

iv. Ek = 1/2mv²
= 40 x (0.02)²/2
= 0.008 J

QUESTION 2

In a harbour, the equation for the depth h of water is h = 5.0 + 3.0sin(2π/45600)
where h is given in metres and t is the time in s. (The angle 2πt/45600 is in radians)
For this harbour, calculate:
i. the maximum depth of water

5.0 + 3.0 = 8.0 m

ii. the minimum depth of water

= 5.0 m

iii. the time interval between high and low-water
iv. two values of t at which the water is 5.0m deep
v. the length of time for each tide during which the depth of water is more than 7.0 m.

I don't understand iii, iv v. May I have some hints on how to approach them please.

 

[Delphi51]

T = 2π√(l/g) = 2π√(8/9.81) = 5.67 s.

I don't follow your work in ii.
You say w = 2πf and f = 0.25 but then you put "w = 2π x 1.25"
You have
x = rsinwt = 2m x sin7.85 1/rads x 0.8
but r is known to be 8 and the time is unknown.
Perhaps knowing the displacement is 2 m, you can find the angle the swing is at and use x = r*sin(θ)

 

[MorallyObtuse]

T = 2π√(l/g) = 2π√(8/9.81) = 5.67 s.

I don't follow your work in ii.
You say w = 2πf and f = 0.25 but then you put "w = 2π x 1.25"
You have
x = rsinwt = 2m x sin7.85 1/rads x 0.8
but r is known to be 8 and the time is unknown.
Perhaps knowing the displacement is 2 m, you can find the angle the swing is at and use x = r*sin(θ)

Thanks for the correction.
Revised calculations

ii. x = rsinwt
w = 2πf
f = 1/5.67
= 0.18 Hz
t = 5.67s
w = 2π x 0.18
= 1.13 1/rads
x = 2m x sin1.13 1/rads x 5.67
= 10.3

iii. V = wrcoswt
= 1.13 1/rads x 2π x cos1.13 1/rads x 5.67
= 17.18 m/s

iv. Ek = 1/2mv²
= 40 x (17.18)²/2
= 5903 J

 

[Delphi51]

x = 2m x sin1.13 1/rads x 5.67

Why are you replacing r with 2 m? Shouldn't it be 8m ?
Why are you using a time of 5.67 seconds? That is the period, the time for one complete swing to one side and back and to the other side and back. The time to swing out 2 m will be much less than 5.67 seconds.

 

[MorallyObtuse]

Why are you replacing r with 2 m? Shouldn't it be 8m ?
Why are you using a time of 5.67 seconds? That is the period, the time for one complete swing to one side and back and to the other side and back. The time to swing out 2 m will be much less than 5.67 seconds.

So, switching the 2m to 8m would make it more proportional?
5.67s was used because it is the only time available from the problem. Also noting that the calculation of time period preceded this question.

x = 8 m x sin1.13 1/rads x 5.67?

 

[Delphi51]

Oops, sorry! I am mistaken and you are right. Maximum displacement of 2 m so the r = 2. I was thinking of the swing going around in a circle.

According to x = rsinwt at time zero the swing is vertical, so it would take time 5.67/4 (a quarter of the period) to swing out to the maximum displacement. Yes, I think that works out to x = 2 m so it all hangs together.

What was the question? Oh yes, the maximum VERTICAL distance, so nothing to do with the frequency or the x = rsinwt. Consider that the swing moves in a circle with radius 8. When it has moved 2 m along that circle, what angle is it at? Hint: it has moved 2 meters out of the full circumference, so it is a fraction of 360 degrees. Once you have that angle you can use trig to get the height.

 

[MorallyObtuse]

Oops, sorry! I am mistaken and you are right. Maximum displacement of 2 m so the r = 2. I was thinking of the swing going around in a circle.

According to x = rsinwt at time zero the swing is vertical, so it would take time 5.67/4 (a quarter of the period) to swing out to the maximum displacement. Yes, I think that works out to x = 2 m so it all hangs together.

What was the question? Oh yes, the maximum VERTICAL distance, so nothing to do with the frequency or the x = rsinwt. Consider that the swing moves in a circle with radius 8. When it has moved 2 m along that circle, what angle is it at? Hint: it has moved 2 meters out of the full circumference, so it is a fraction of 360 degrees. Once you have that angle you can use trig to get the height.

l = 2 m
= 2/2π
= 0.32

 

[Delphi51]

2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees

 

[MorallyObtuse]

2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees

2π x 8 = 50.3
50.3 - 2 = 48.3?

 

[Delphi51]

2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees
2/(2*pi*8) = A/360
Solve for A in degrees.

 

[MorallyObtuse]

2 meters out of a circumference of 2*pi*8 = angle out of 360 degrees
2/(2*pi*8) = A/360
Solve for A in degrees.

2/(2π x 8) = A/360
720/(2π x 8) = A
A= 720/(2π x 8)
= 14.32

I'm not sure of how to calculate the height, considering that theta = 14.32, r = 8

 

[MorallyObtuse]

Question 2 should read

In a harbour, the equation for the depth h of water is h = 5.0 + 3.0 sin (2π t/45600)
where h is given in metres and t is the time in s. (The angle 2πt/45600 is in radians)
For this harbour, calculate:

 

[Delphi51]

If you can find the side "a", then you can get "h" easily.

#2: Better check the minimum. Suggest you graph the sine function on your calculator so you can see how it goes. You can get a good estimate of the time interval between high and low that way, too. If you calculate the period of the sinusoidal, you can easily get the exact value for that interval. In sin(kt) the period is 2π/k.

 

[MorallyObtuse]

If you can find the side "a", then you can get "h" easily.

#2: Better check the minimum. Suggest you graph the sine function on your calculator so you can see how it goes. You can get a good estimate of the time interval between high and low that way, too. If you calculate the period of the sinusoidal, you can easily get the exact value for that interval. In sin(kt) the period is 2π/k.

sin14.32 = x/8
x = 8sin14.32
= 1.98 m

(8)² = a² + (1.98)²
64 - 3.92 = a²
60.08 = a²
a = √60.08
= 7.75 m

x = 1.98 m; 2 m =hyp; h =?
2² = 1.98² + (hyp)²
4 - 3.92 = hyp²
0.08 = (hyp)²
hyp = √0.08
= 0.28 m; Hence h = 0.28 m

I'm not understanding this.

t = 0 is the minimum depth of water

2πt/45600 = π/2
t = 45600π/ 4π
= 11 400 s (low water)

2πt/45600 = π
t = 45600/2π
= 22 800 s (high water)

Time interval between high and low water: 22 800 - 11 400
= 11 400 s

depth of water (l) = 8.0 m
g = 9.8 m/s ²
T= ?
Eqn. for T = 2π√(l/g)
= 2π√8/9.8
= 1.81 s

 

[Delphi51]

You have side a = 7.75 correct. To find the height, just subtract
h = 8 - 7.75 = 0.25.

If you graph #2, you will see:

In sin(kt) the period is 2π/k. Your k = 2π/45600 so you have
period = 2π/(2π/45600) = 45600. The time from one crest to the next is the period, 45600 seconds. Think about where the minimum is compared to the two crests. Check your answer by looking at the graph.

 

[MorallyObtuse]

You have side a = 7.75 correct. To find the height, just subtract
h = 8 - 7.75 = 0.25.

If you graph #2, you will see:

In sin(kt) the period is 2π/k. Your k = 2π/45600 so you have
period = 2π/(2π/45600) = 45600. The time from one crest to the next is the period, 45600 seconds. Think about where the minimum is compared to the two crests. Check your answer by looking at the graph.

I don't get 45600 when I subtract a 1000 (initial) from 35000 (minimum point) - the difference between the two crests. Explain please.
In sin(kt) the period is 2π/k. Your k = 2π/45600 so you have
period = 2π/(2π/45600) = 45600. I don't understand the equations you put forth?!

 

[Delphi51]

The first 2 crests are at about 11000 and 57000.
A crest is a maximum, the top of the wave.

 

[MorallyObtuse]

The first 2 crests are at about 11000 and 57000.
A crest is a maximum, the top of the wave.

Is the minimum = 35000?
So, how do I go about obtaining the time interval between high and low-water?

 

[Delphi51]

Yes, roughly 35000. 34000 would be better.
So, you've got a low at 34000 and a high at 11000.
Just subtract to get the time interval between them - approximately.
To get an accurate answer you must look again at the wave formula and calculate the period. Use that T = 2π/k trick I mentioned earlier.

 

[MorallyObtuse]

Yes, roughly 35000. 34000 would be better.
So, you've got a low at 34000 and a high at 11000.
Just subtract to get the time interval between them - approximately.
To get an accurate answer you must look again at the wave formula and calculate the period. Use that T = 2π/k trick I mentioned earlier.

I'm not sure what you mean by accurate here. I get answers at different ends of the spectrum:)

T = 2π/(2π/45600) = 45 600
Time interval = 34 000 - 11 000 = 23 000

iv. two values of t at which the water is 5.0m deep

(0,0) seeing that these are the coordinates

v. the length of time for each tide during which the depth of water is more than 7.0 m.

Readings from the graph.
Tide 1 app. 7 000s Tide 2 is app. 50 000

 

[Delphi51]

T = 2π/(2π/45600) = 45 600
Time interval = 34 000 - 11 000 = 23 000

Good work! The 45 600 is the period, the time from high to high.
The 23 000 is an estimate of the time from high to low.
Do you see a relationship between the two? Look on the graph - how does the time from high to low compare with the time from high to high?
Still in the dark - adjust your estimate from 23000 to 22800 and try again.

 

[MorallyObtuse]

Good work! The 45 600 is the period, the time from high to high.
The 23 000 is an estimate of the time from high to low.
Do you see a relationship between the two? Look on the graph - how does the time from high to low compare with the time from high to high?
Still in the dark - adjust your estimate from 23000 to 22800 and try again.

Yes, the time from high to low is half that of the time from high to high.

 

[Delphi51]

All right! Take half of the accurate value you have for the period.

 

[MorallyObtuse]

All right! Take half of the accurate value you have for the period.

Thanks very much for your patience and help

 

[Delphi51]

Most welcome!

 

[MorallyObtuse]

# 1
Check this also, please.

iii. Maximum velocity of the child during the motion
Maximum velocity (Vm) = wr
f = 1/T = 1/5.67 = 0.18 Hz
Now w = 2πf = 2π x 0.18, and r = 2.0 m
therefore Vm = 2π x 0.18 x 2.0 = 2.26 m/s

iv. The total energy produced due to this motion
Total energy (Ek) = 1/2 mw²r²
= 1/2 x 40 x (2π0.18)² x (2.0)²
= 20 x 1.28 x 4
= 102.4 J

 

[Delphi51]

Looks good! I did get a slightly different answer when I did not round off in middle steps.
V = rω = r2πf = 2πr/T = 2π*2/5.674 = 2.21
If you want 3 digit accuracy in your final answer, you must keep 4 digit accuracy in the middle steps.

 

[MorallyObtuse]

Looks good! I did get a slightly different answer when I did not round off in middle steps.
V = rω = r2πf = 2πr/T = 2π*2/5.674 = 2.21
If you want 3 digit accuracy in your final answer, you must keep 4 digit accuracy in the middle steps.

Thanks once more