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What factors affect the rate at which my spring Oscillates?

Extracts from this document...


Clare Heirene 11I


Contents                                                                        Page 1

Brainstorm                                                                        Page 2

Hypothesis                                                                        Page 3


        Diagram and List Of Apparatus                                        Page 5        Method                                                                Page 6

        Fair Test                                                                Page 7

        Safety                                                                        Page 8


        Table Of Results                                                        Page 9

        Graph                                                                        Page 10

Interpretation                                                                Page 13

Evaluation                                                                        Page 14

Appendix                                                                        Page 15



I think that the more weight you put on the spring, the more time it will take to make ten oscillations. To explain this I will start off at the beginning. According to Hooke’s Law1 the extension is directly proportional to the force loaded onto it. The graph below shows the load and the extension. The line is straight meaning x=y or extension α load.


 I think that if you doubled the weight you added to the spring then the extension of the spring will double as well2.

I think that this will only happen, though, up until a certain point, because after that the spring will not revert to its original shape3. This point is called the elastic limit. The graph below shows the elastic limit of a spring. As you can see, at a certain point Hooke’s law fails to work4. It starts off as E α L until it reaches the elastic limit where the spring extends more than it would if E α L. This is where the spring starts to stretch out of shape and will not go back to it’s original state.


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I will put masses of 100g onto the spring and add 100g each time. I will take a set of 5 results from my experiment. I will then repeat the experiment twice to find repeat readings and averages. I will take readings of how long, in seconds, it takes the spring to oscillate ten times. One oscillation is both an up and a down motion. I will give each a pull of 8cms.

To perform my experiment I will follow these points:

  1. Set up the apparatus as shown in the diagram.
  2. Load a weight onto the hanger.
  3. Use the ruler to measure how far you pull the spring.
  4. Let go of the spring and start the timer at the same time.
  5. Count ten bounces and stop the timer.
  6. Record the result in a table.
  7. Repeat steps 2-6 adding 1 more weight each time you repeat them.
  8. Repeat the experiment twice.

I will record my results in these tables:

Mass (g)

Time taken for 10 Oscillations (s)

2nd Experiment Results (s)

3rd Experiment Results (s)






Mass (g)

Averages (s)






Fair Test

To make my experiment fair, I will follow these guidelines:

  • Make sure you pull the spring down the same amount for each result.
  • Make sure you use the same equipment for each experiment, as they could be slightly different.
  • Be as accurate as possible. Start the clock as soon as you let go of the spring and stop it as soon as it has oscillated ten times.
...read more.


l as in the figure below (a), then if k is the tension required to produce unit extension (called the spring constant and measured in N m-1) the stretching tension is also kl and so



Suppose the mass is now pulled down a further distance x below its equilibrium position, the stretching tension action downwards is k (l+x) which is also the tension in the spring acting upwards, this is shown in the picture above (b). Hence the resultant restoring force upwards on the mass

= k (l+x) – mg

   = kl + kx – kl    (since mg=kl)

      = kx

When the mass is released it oscillates up and down. If it has an acceleration a at extension x then by Newton’s second law

-kx = ma

The negative sign indicates that at the instant shown a is upwards while the displacement x is downwards.


 a= - m  x = - ω2x

where ω2 =k/m = a positive constant since k and m are fixed. The motion is therefore simple harmonic about the equilibrium position so long as Hooke’s law is obeyed. The period T is given by 2π /ω, therefore

T= 2πm


It follows that T2 = 4π2 m/k. If the mass m is varied and the corresponding periods T found, a graph of T2 against m is a straight line but it does not pass through the origin as we might expect from the above equation. This is due to the mass of the spring itself being neglected in the above derivation. Its effective mass and a value of g can be found experimentally.

(Advanced Physics Fourth Edition by Tom Duncan. Page 178 Mass on a spring. (a)Period of Oscillations.)

        -  -

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