# Spring constant

1. Jan 25, 2010

### zhenyazh

hi,
an image is attached.

The left side of the figure shows a light (`massless') spring of length 0.260 m in its relaxed position. It is compressed to 78.0 percent of its relaxed length, and a mass M= 0.150 kg is placed on top and released from rest (shown on the right).

The mass then travels vertically and it takes 1.30 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

energy conservation cannot work here because i don't have enough info.
thanks

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2. Jan 25, 2010

### D H

Staff Emeritus
What information are you lacking? (Hint: You do have adequate information here.)

3. Jan 25, 2010

### zhenyazh

well i think i lack the final energy.
i guess i can use the information about time to help me but i don't know how.

4. Feb 6, 2010

### zhenyazh

anyone?

i have the spring energy at first and i can calculate the potential
energy when the mass is at the top. the problem is k is both in cos and outside it

5. Feb 6, 2010

### D H

Staff Emeritus
There are no non-conservative forces in this problem, so the total energy (kinetic+potential) is constant. The velocity is zero initially and at the top of the trajectory: The energy is purely potential at these points.

How high does the object go in 1.3 seconds? What does that mean in terms of potential energy, and how does that relate back to the spring constant?

6. Feb 6, 2010

### zhenyazh

this is exactly the point.
to find the height relative to the spring rest position
i need to use the harmonic motion x function.
but some how it goes wrong.
first i am not sure i am assigning values correctly
and second when i do it i have the constant k both inside the cos and outside

7. Feb 6, 2010

### D H

Staff Emeritus
You don't need a harmonic motion function. The mass is resting on the spring, not attached to it. The mass flies off of the spring at some point.

8. Feb 6, 2010

### D H

Staff Emeritus
Correction to the above:

I see what your concern is, zhenyazh. You don't need to worry about the harmonic motion only if that 1.3 seconds is measured from the point where the mass flies off the spring. You do need to worry about the motion while the mass is on the spring if that 1.3 seconds is measured from the time when the process is started. It is not clear what sense is meant from the wording in the original post.

Suppose it is the latter. The time of flight is 1.3 seconds less the time it takes the spring to fling the mass upward, and that time spent on the spring depends on the spring constant. It would seem you have a chicken and egg problem. You need the solution in order to find the solution.

That isn't quite the case here. Suppose you have an initial guess regarding the solution. From this you can calculate the time spent on the spring. Subtracting that from 1.3 seconds gives the time of flight. The time of flight gives a new final energy, and from that you can calculate an improved estimate of the spring constant. You could iterate here, but that won't be necessary. Assuming you start with a reasonably good guess, the first iteration will remove almost all of the error.

So what to use for an initial guess? Simple: Ignore the time spent on the spring.time of flight you can refine that guess.