SHM: Different Final Answers When Using Energy Conversation & Trigonometry

[Oz Alikhan]

1. The Question from Text:

"A forest playground has a tyre hanging from a tree branch. The tyre behaves like a pendulum, with a rope of 4.0 metres length, and the tyres mass 15 kg. A child of mass 45 kg swings on the tyre by pulling it 3.0 metres to one side and leaping on. What is the maximum height that the tyre will reach above its equilibrium position?" 3. The Solutions:

Through Energy Conservation

T = 2∏ \sqrt{l/g} ≈4.0s
ω = 2∏/T ≈1.6 rad/s

Therefore, V_max = Aω = 3.0 x 1.6 = 4.7 (Using unrounded T & ω)

Max K.E = 0.5*(15+45)*(4.7)^2 = 660 J

Max P.E = 660 J = mgh
h = 660/mg = 1.1 mThrough Trigonometry

The right angled triangle: Cosθ = 3/4
If I use Pythagoras Theorem, I get \sqrt{4^{2} - 3^{2}} = \sqrt{7}
Then, 4 - \sqrt{7} = 1.35 mWhy is there a significant difference in my answers? (Quick side note: Why doesn't my MathTex not work?)

 

[gneill]

1. The Question from Text:

"A forest playground has a tyre hanging from a tree branch. The tyre behaves like a pendulum, with a rope of 4.0 metres length, and the tyres mass 15 kg. A child of mass 45 kg swings on the tyre by pulling it 3.0 metres to one side and leaping on. What is the maximum height that the tyre will reach above its equilibrium position?"


3. The Solutions:

Through Energy Conservation

T = 2∏ \sqrt{l/g} ≈4.0s
ω = 2∏/T ≈1.6 rad/s

Therefore, V_max = Aω = 3.0 x 1.6 = 4.7 (Using unrounded T & ω)

Max K.E = 0.5*(15+45)*(4.7)^2 = 660 J

Max P.E = 660 J = mgh
h = 660/mg = 1.1 m


Through Trigonometry

The right angled triangle: Cosθ = 3/4
If I use Pythagoras Theorem, I get \sqrt{4^{2} - 3^{2}} = \sqrt{7}
Then, 4 - \sqrt{7} = 1.35 m


Why is there a significant difference in my answers? (Quick side note: Why doesn't my MathTex not work?)

A pendulum is not a "good" SHM device except for small displacement angles (where "small" means sin(θ) ≈ θ). Here your angle is nearly 50°, so expect large inaccuracies!

MathTex expressions should be between appropriate tags that flag MathTex to interpret the enclosed text. For expression embedded in text you can use [ itex ] and [ /itex ] (no spaces) or a pair of ## . Larger, one expression per line version uses tags[ tex ] ... [ /tex ] (again no spaces) or a pair of $$'s

 

[Oz Alikhan]

Thanks for the reply. That is what I thought, although I tried to improve this approximation by using the formula that takes into account bigger amplitudes: http://upload.wikimedia.org/wikipedia/en/math/3/0/1/3014cdccf9d3df9770a001b84ae6ac80.png

However, the percentage error difference between my two answers even got bigger after I used that formula.

 

[Oz Alikhan]

Oh I see. The Trigonometry method is a linear approximation that does not take into account of the larger angles where the approximation deviates from this linearity. That explains the difference in answers .

Thanks for clearing that up