Child oscillating on playground swing

In summary, the conversation discusses a homework problem involving a child on a playground swing and their period of oscillation. The first part of the problem is solved correctly using the formula l = (980)(2.5)^2/4n^2, and the length of the swing is calculated to be 122.93 cm. However, when a second child with twice the weight is added to the swing, the period of oscillation is asked for. The conversation includes a hint that the mass does not affect the period, and the formula T = 2 pi sqrt(L/g) is mentioned. The conversation also briefly touches on the concept of mass not affecting the period in general, and how this relates to the formula for the period
  • #1
brayrbob
24
0
Here is a homework problem that I believe is solved right.

A child oscillates freely on a playground swing. It takes 2.5 seconds for the child to complete one full swing and return to the person that is pushing her. How long is the swing? This is how I solved for the answer.

l = (980)(2.5)^2/4n^2 = 15112.83174 (the n stands for pi)
then I took the square root of 15112.83174 to get 122.9342578 or 122.93 cm

The next part of the problem I don't know what to do.
A child twice as heavy sits on the same swing. What is his period of oscillation? How do you Know?
I'm not real sure where I'm supposed to start with this problem.
 
Physics news on Phys.org
  • #2
In the first part you're pretty much right, but it should say [itex]l^2=[/itex] instead of l=, and the first term should be [itex]\sqrt{g}[/tex] intead of g. For the second part, a hint: does the mass appear in the equation for the period? What does that tell you?
 
  • #3
Hmmm not totally sure about that hint.
The period is the time so 2.5^2 is the period of time, but I don't see where the mass fits in. Do you divide how long the swing is by the time.
122.93/2.5 = 49.172 to get the mass and then double that to get 99.42?
 
  • #4
the hint is, mass doesn't make a difference...
 
  • #5
So, that means that mass doesn't make a difference in this particular problem only.
Wouldn't a heavier child make the length of the swing change, or is this question asking for the length of the swing not the length of the oscillation?
 
  • #6
The mass of the kid doesn't affect the period of oscilation. That means that if they gave you the period of oscillation and asked for the length of the swing, the kid's mass would not be a factor. This is the case here because gravity is the force and the force of gravity is proportional to the mass. This causes a cancelation in the differential equation from which the formula for the period is derived. If we were talking about a block on a spring lying on the table, then the mass would change the period.
 
  • #7
Since gravitational force is proportional to m and F= ma, acceleration, and therefore movement, due to gravity, in general, does not depend on the mass of the accelerating body. For a pendulum, which is essentially what you have here, period depends only on the square root of the length of the pendulum.
 
  • #8
I understand this problem, but I tried to work another problem in the same manner and my professer said I did it the wrong way. Here is that problem and the correct way to work it.
How long should the lenth of a pendulum be to have a period of exactly 1.000 seconds on a "grandfather" clock?
l=[(980)(1.000)^2]/(4pi^2) = 24.82368999 or 24.82cm
My question is what is the basic difference between these two questions and why wouldn't I take the square root of the answer in this question?
 
  • #9
Do you understand the difference between "square" and "square root"? What I said was that the period is proportional to the square root of the length. That is,[tex]T= k\sqrt{l}[/tex] (and the k is, as you gave, [tex]\frac{2\pi}{\sqrt{g}}[/tex]) so [tex]T^2= k^2l[/tex] and [tex]l= \frac{T^2}{k^2}[/tex], exactly the formula you give. The length is proportional to the square of the period.
 
Last edited by a moderator:
  • #10
Square is when I square the number like 2^2 and square root is when I take the square root of the answer. I'm in intro to Physics and doing the lab part and I understand some of the basics but tend to have problems of knowing what to do and when to do it. I'm trying hard to understand all of it but the equations give me problems, especially since every problem has a new formula to use.
 
  • #11
brayrbob said:
I understand this problem, but I tried to work another problem in the same manner and my professer said I did it the wrong way. Here is that problem and the correct way to work it.
How long should the lenth of a pendulum be to have a period of exactly 1.000 seconds on a "grandfather" clock?
l=[(980)(1.000)^2]/(4pi^2) = 24.82368999 or 24.82cm
My question is what is the basic difference between these two questions and why wouldn't I take the square root of the answer in this question?
The two problems are essentially identical. This one is done correctly. The one you did in your first post was done incorrectly: (1) You made an arithmetic mistake, (2) for some reason, you took the square root.

The period of a simple pendulum is:
[tex]T = 2 \pi \sqrt{\frac{L}{g}}[/tex]

Solving for the length:
[tex]L = \frac{g T^2}{4 \pi^2}[/tex]
 
  • #12
I have trouble remembering when to take the square root of something. You all are very helpful but algebra and physics are hard for me to understand. I do understand the basics but trust me I will be glad when I'm through with these courses. It's like a foreign language to me that I have to force myself to concentrate on.
I did rework the first problem l=((980)(2.5)^2)/(4pi^2)) = 155.15cm
 

1. What is the science behind a child oscillating on a playground swing?

The science behind a child oscillating on a playground swing involves the principles of physics, specifically the concept of pendulums. The child's weight and movements create a pendulum motion, where the swing moves back and forth due to gravity and the child's force.

2. Why do children enjoy swinging on playground swings?

Children enjoy swinging on playground swings because it provides a sense of thrill and excitement. The swinging motion also stimulates the vestibular system, which helps with balance and coordination, making it a fun activity for children.

3. Is swinging on a playground swing beneficial for a child's development?

Yes, swinging on a playground swing can be beneficial for a child's development. It helps with gross motor skills, balance, and coordination. It also provides sensory input and can help with emotional regulation and relaxation.

4. How can a child make the swing go higher?

A child can make the swing go higher by pumping their legs back and forth while leaning back. This motion increases the swing's momentum and can make it go higher. The length of the swing's chains also plays a role in how high it can go.

5. Are there any safety concerns when a child is swinging on a playground swing?

Yes, there are some safety concerns when a child is swinging on a playground swing. It is important to make sure the child is supervised and that they are using a swing appropriate for their age and size. It is also essential to check the swing's condition and ensure it is stable and properly anchored to the ground.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
19
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
7K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
30
Views
2K
Back
Top