# Homework Help: Spring Question

1. Jun 22, 2008

### zooboodoo

1. The problem statement, all variables and given/known data
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. Relevant 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 dont 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

2. Jun 22, 2008

### 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?

3. Jun 22, 2008

### 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?

4. Jun 22, 2008

### zooboodoo

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

5. Jun 22, 2008

### rock.freak667

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?

6. Jun 22, 2008

### 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)

7. Jun 22, 2008

### rock.freak667

Correct and 12.566 is the angular frequency.

Now for a spring,$\omega=?$

8. Jun 22, 2008

### 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?

9. Jun 22, 2008

### rock.freak667

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

10. Jun 22, 2008

### rock.freak667

Yep, should be correct.

11. Jun 22, 2008

### 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

12. Jun 22, 2008

### rock.freak667

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

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

13. Jun 22, 2008

### 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

14. Jun 22, 2008

### rock.freak667

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.

15. Jun 22, 2008

### 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 in to maximum KE=1/2 mA^2 * w^2, solve for w?

16. Jun 22, 2008

### rock.freak667

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)

17. Jun 22, 2008

### zooboodoo

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