Solving Spring Constant: x(t) = 1.5*sin(12.566*t)

[zooboodoo]

1. Homework Statement
A 300 gram mass is attached to a massless spring and allowed to oscillate around an equilibrium according to the equation: x(t) = 1.5*sin(12.566*t). where x is measured in meteres and t in seconds.

a.) What is the spring constant
b.) What is the total mechanical energy of the system
c.) what is the maximum kinetic energy of the mass
d.) what is the maximum velocity of the sytem



2. Homework Equations
F = -kx
Usp= (1/2)kx^2

I think the problem I am having is attempting to relate enough variables in one equation to try to solve for the spring constant, I think I understand that B and C will be the same, when Kinetic energy is at the highest potential energy will be = 0 and that will be same as the total mechanical energy, I just feel like I'm lost trying to solve for the spring constant



3. The Attempt at a Solution
I thought of trying to graph the equation of the graph given however i don't think that this is the apporpriate method to solve the problem, also i thought about substituting an arbitrary time to and try to solve for final velocity working backwards however I'm not sure how to use the equations I've learned to solve for these variables

 

[rock.freak667]

In the equation x(t) = 1.5*sin(12.566*t), do you know what 1.5 represents and what the 12.566 represents?

 

[zooboodoo]

I found another equation online, x=Acos((2(pi)/T) * t) this is position versus time in simple harmonic motion which means that 1.5 is the factor by which it should oscillate between A/-A now from this can i calculate the maximum velocity by using simple substitution?

 

[zooboodoo]

I believe the 1.5 would be the amplitude, and the 12.66 would be 2pi / T = 2pi/frequency?

 

[rock.freak667]

I found another equation online, x=Acos((2(pi)/T) * t) this is position versus time in simple harmonic motion which means that 1.5 is the factor by which it should oscillate between A/-A now from this can i calculate the maximum velocity by using simple substitution?

Yes,that is another equation for SHM when you start at max displacement.

But similarly, the equation you have is in the form x=Asin \omega t

A is the amplitude and \omega is the angular frequency.

For a spring how does \omega relate to the spring constant,k?

 

[zooboodoo]

i think i solved for T, 2pi / T = 12.566, T = .500, the equation relating w and k would be, w=(k/m)^.5 ,err, w= root(k/m)

 

[rock.freak667]

I believe the 1.5 would be the amplitude, and the 12.66 would be 2pi / T = 2pi/frequency?

Correct and 12.566 is the angular frequency.

Now for a spring,\omega=?

 

[zooboodoo]

ohhh so if 12.566 is the angular velocity, 12.566=root(k/m) k = 12.566^2 * m (in kilograms?) , k = 47.37?

 

[rock.freak667]

i think i solved for T, 2pi / T = 12.566, T = .500, the equation relating w and k would be, w=(k/m)^.5 ,err, w= root(k/m)

right, w=root (k/m) and w=12.566 from your equation.

 

[rock.freak667]

ohhh so if 12.566 is the angular velocity, 12.566=root(k/m) k = 12.566^2 * m (in kilograms?) , k = 47.37?

Yep, should be correct.

 

[zooboodoo]

thank you very much! is it safe to replace frequency with "angular velocity" in other cases, for example when i try to solve for the mechanical energy

 

[rock.freak667]

thank you very much! is it safe to replace frequency with "angular velocity" in other cases, for example when i try to solve for the mechanical energy

That was a typo on my part, the \omega is angular frequency.

Do you know the formula for the total energy when undergoing SHM?

 

[zooboodoo]

E=Umax=1/2k(A)^2 , so, (1/2)(47.37)(1.5)^2 = E, = 53.29 J, does this seem like an oversimplification, or will my mass show up again, aside from in the calculation for k

 

[rock.freak667]

E=Umax=1/2k(A)^2 , so, (1/2)(47.37)(1.5)^2 = E, = 53.29 J, does this seem like an oversimplification, or will my mass show up again, aside from in the calculation for k

Correct.

You'll need to use the mass for another part.

and you know the answer to the other part for the max. kinetic energy.

 

[zooboodoo]

I'm sorry I got slightly confused there, the equation is correct, but I need to use the mass again? I believe this was my total mechanical energy, which should be the same as the maximum kinetic energy, and from there, to solve the maximum velocity, i can plug into maximum KE=1/2 mA^2 * w^2, solve for w?

 

[rock.freak667]

I'm sorry I got slightly confused there, the equation is correct, but I need to use the mass again? I believe this was my total mechanical energy, which should be the same as the maximum kinetic energy, and from there, to solve the maximum velocity, i can plug into maximum KE=1/2 mA^2 * w^2, solve for w?

The max kinetic energy is 53.29 J, and KE=1/2 mv^2. Solve for v. (Not sure if that is what you meant but I took 'w' to mean angular frequency)

 

[zooboodoo]

right, i got 53.29 and then used 1/2mv ^2 to get 18.85 for maximum velocity, thank you very much