Simple harmonic motion energy question

AI Thread Summary
The discussion revolves around solving a homework problem related to simple harmonic motion and energy conservation. Participants emphasize that total energy, the sum of kinetic energy (KE) and potential energy (PE), remains constant. When KE is three-quarters of the total energy, the remaining fraction is identified as potential energy. The potential energy of a spring is defined as PE = 1/2 kx², derived from Hooke's law. The conversation also touches on alternative methods for solving the problem, including trigonometric and calculus approaches, while addressing the participant's confusion about the placement of variables in the equations.
Samurai44
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Homework Statement


Can I get help in part (b) of this question ?
Q1.jpg


Homework Equations


KE=1/2 m v2
v= (2π f )√(A2 - x2)

The Attempt at a Solution


I substituted the second equation into first one, so i got
KE= 1/2 m (2π f )2 (A2 - x2)

but then couldn't complete
 
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You are apparently concerned with the answer to part b. One approach starts with the insight that total energy is conserved. The sum of PE + KE is a constant.

If KE is 3/4 of total energy then what fraction of total energy is PE?

The next insight is that the force required for harmonic motion is exactly the force supplied by a perfect spring. Or, simpler yet, we could simply assume that the oscillation is produced by a mass attached to a perfect spring. Do you know the formula for the potential energy of a spring?
 
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jbriggs444 said:
You are apparently concerned with the answer to part b. One approach starts with the insight that total energy is conserved. The sum of PE + KE is a constant.

If KE is 3/4 of total energy then what fraction of total energy is PE?

The next insight is that the force required for harmonic motion is exactly the force supplied by a perfect spring. Or, simpler yet, we could simply assume that the oscillation is produced by a mass attached to a perfect spring. Do you know the formula for the potential energy of a spring?
Do you mean hooks law F=-kx ?
 
Yes. That's part of it. There is also a formula for energy in a spring as a function of displacement x. That formula can be derived by integrating F=-kx over distance. The result is PE = 1/2kx2
 
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Hi,
You know that the energy is conserved right, so E is constant, why not calculating is at any point, example would be when there's no kinetic energy, Once you got E, you know that when KE = 3E/4, what is left if energy is potential energy, do this sound familier kx^2 /2 ?
 
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jbriggs444 said:
Yes. That's part of it. There is also a formula for energy in a spring as a function of displacement x. That formula can be derived by integrating F=-kx over distance. The result is PE = 1/2kx2
Noctisdark said:
Hi,
You know that the energy is conserved right, so E is constant, why not calculating is at any point, example would be when there's no kinetic energy, Once you got E, you know that when KE = 3E/4, what is left if energy is potential energy, do this sound familier kx^2 /2 ?
well , iam not used to PE = 1/2kx2 :(
 
Kx^2/2 is the elastic potential energy of a system, where K is the spring's constant, x is the strech of the spring, and 1/2 is one half (obviously, ) and if you want a proof, W = F.x so dW = F*dx = F*x'dt = k*x*x' dt and you integrate both sided and it yields to W = 1/2kx^2 ! And now you can work with it, good luck
 
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Samurai44 said:
well , iam not used to PE = 1/2kx2 :(

If you don't like using PE = 1/2 kx2 there are still a couple of other approaches to solving the original problem. Do you prefer trigonometry or differential calculus?
 
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jbriggs444 said:
If you don't like using PE = 1/2 kx2 there are still a couple of other approaches to solving the original problem. Do you prefer trigonometry or differential calculus?
Isnt it possible to solve it by the equation I have given ?
it will be the same on both sides , but one will have the coefficient (3/4).
At the same time , it say in equilibrium position , so one side will have x=0, and so max. velocity (K.E)
but i am facing problem in which side to place them ( right or left ).
correct me if I am wrong please .
 
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Samurai44 said:
Isnt it possible to solve it by the equation I have given ?
it will be the same on both sides , but one will have the coefficient (3/4).
At the same time , it say in equilibrium position , so one side will have x=0, and so max. velocity (K.E)
but i am facing problem in which side to place them ( right or left ).
correct me if I am wrong please .

The equation seems to be missing a factor of sin(theta).

Edit: I take that back. That's what the ##\sqrt{A^2-x^2}## is supposed to cover.

So if you know that the total energy E is equal to KE alone when x=0 (because this is the point where PE is presumably taken as zero), that should give you an equation.
 
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