• Support PF! Buy your school textbooks, materials and every day products Here!

SHM child on a swing question

  • Thread starter blackcat
  • Start date
  • #1
57
0
Hi,

A child on a swing swings with a time period of 2.5s and an amplitude of 2m.

What is the max. kinetic energy of the oscillation?

I'm not sure how to work this out without her mass. Her max speed is 2.51m/s but I dunno how to do this. BTW this is all the information that is given

Any hints?
 
Last edited:

Answers and Replies

  • #2
Hint: Draw a picture and use the formula for period of a pendulum. Once you find the height between the highest and lowest points, you can find the kinetic energy.
 
  • #3
andrevdh
Homework Helper
2,128
116
During the SHM motion the oscillator continually converts potential energy to kinetic energy and back. At the extremes of its motion it momentarily comes to rest. At these points all energy is converted to potential energy. When the oscillator is at its equilibrium position all of its energy is converted back to kinetic energy. So try and find the maximum potential eneregy of the oscillator. It might be helpfull to totally forget that you are dealing with a swing and just concentrate on the maths. Anyway, the statement that the amplitude of the swing is 2 meters can be interpreted in many ways.
 
  • #4
57
0
Ok thanks both of you.
 
  • #5
andrevdh
Homework Helper
2,128
116
What I was trying to say is that the maximum kinetic energy of the oscillator should be equal to

[tex]\frac{1}{2}kA^2[/tex]

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" [Broken]

so all you need is the "spring constant for the pendulum" - which unfortunately does depend on the mass!

http://theory.uwinnipeg.ca/physics/shm/node5.html" [Broken]
 
Last edited by a moderator:
  • #6
andrevdh
Homework Helper
2,128
116
Looking at it differently one can say

[tex] \Gamma = I \alpha [/tex]

which gives

[tex] \ddot{\theta} = \frac{1}{I} \Gamma[/tex]

for

[tex]\Gamma = lw \sin(\theta) [/tex]

for small swing angles (which the condition for SHM for a pendulum) one gets

[tex] \Gamma = lmg\theta [/tex]

which gives the more prommising (maybe?) SHM equation

[tex] \ddot{\theta} = -\frac{g}{l} \theta [/tex]

the justification for inserting the - is that the torque is positive (anticlockwise) when the angle is negative (to the left of the equilibrium) and vice versa when the pendulum is on the other side of the equilibrium.

But I do'nt think that one can get pass the fact that the total energy of a pendulum do depend on the mass. For the spring not so. This can be understood on the basis that the energy is stored in totallity in the spring when it is strecthed (compressed) to its max, but for the pendulum the max energy depends on the mass swinging from it. If a larger mass swings up to the same height on the same length of string the total energy of the system will just be more. And as we all know the period does not depend on the mass, just the length.
 
Last edited:

Related Threads for: SHM child on a swing question

  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
13
Views
19K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
955
Replies
6
Views
6K
Replies
11
Views
2K
Top