# SHM: Different Final Answers When Using Energy Conversation & Trigonometry

• Oz Alikhan
In summary, the tyre will reach its maximum height when it is 3.0 metres from the equilibrium position, or 4.7 metres using Energy Conservation or 1.35 metres using 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

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?)

Last edited:
Oz Alikhan said:
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

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

1 person
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

I would like to point out that both solutions are correct and valid, but they are using different methods to calculate the maximum height of the tyre. The first solution uses the principle of energy conservation, which considers the initial and final energy of the system. The second solution uses trigonometry to calculate the height of the tyre based on the angle at which the child pulls the tyre.

The difference in the answers can be attributed to the fact that the two methods are based on different assumptions and equations. The energy conservation method assumes that all the energy is converted from kinetic to potential at the highest point, while the trigonometry method assumes that the height of the tyre at the highest point is equal to the vertical displacement of the child.

It is important to note that both methods have their own limitations and may not give the exact same answer. It is also worth mentioning that the difference in the answers may be due to rounding in the calculations.

As for the issue with MathTex, it is likely a technical issue and I would suggest seeking help from a technical support team. In the future, it would be helpful to provide more context and information about the issue to better address it.

## 1. What is SHM?

SHM stands for Simple Harmonic Motion. It refers to the back and forth motion of an object around a fixed equilibrium point, where the force acting on the object is directly proportional to its displacement from the equilibrium point.

## 2. How is energy conservation used in SHM?

Energy conservation is used in SHM to explain how the total energy of the system (kinetic and potential) remains constant throughout the motion. As the object moves back and forth, the energy is constantly being converted between kinetic and potential energy, but the total amount remains the same.

## 3. How is trigonometry used in SHM?

Trigonometry is used in SHM to find the displacement, velocity, and acceleration of the object at any given point in time. The displacement can be represented by a sine or cosine function, while the velocity and acceleration can be found by taking the first and second derivatives of the displacement function.

## 4. What are the different final answers when using energy conservation and trigonometry in SHM?

The final answers when using energy conservation and trigonometry in SHM may differ depending on the specific problem and variables involved. However, they will always be related to the displacement, velocity, and acceleration of the object at a particular point in time.

## 5. How is SHM used in real-life applications?

SHM is used in many real-life applications, including pendulum clocks, musical instruments, and car suspensions. It is also used in seismology to study earthquakes and in engineering to design structures that can withstand vibrations and oscillations.

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