Simple harmonic motion energy question

[Samurai44]

Homework Statement

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

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

 

[jbriggs444]

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?

 

[Samurai44]

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 ?

 

[jbriggs444]

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]

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 ?

 

[Samurai44]

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

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 :(

 

[Noctisdark]

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

 

[jbriggs444]

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?

 

[Samurai44]

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 .

 

[jbriggs444]

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.