Question

For a linear spring the force is proportional to the stretch. F =kx k is the elastic constant of the spring in N/m

Masses are placed on a spring and the stretch is measured (no bouncing). Then the spring was allowed to bounce and the period was measured.

mass Stretch(cm) Period (s)

0.15 kg 9 0.6

0.25 15 0.76

0.35 20.5 0.9

0.45 26.5 1.0

Calculate k for the spring. Be careful mass must be converted to weight in Newtons and the stretch must be in meters.

Plot a graph of period
squared (y) vs. mass. Calculate the slope. The slope is 4
pi^{2} / k. Calculate k from the slope.

Get the % difference between the two k's.

Answer #1

Mass |
Stretch (cm) =X | Force = massX g(9.8) Force = Kx |
K N/m |

.15 | 9 | 16.33 | |

.25 | 15 | 16.33 | |

.35 |
20.5 | 16.73 |

.45 | 26.5 | 16.64 | |

Average | 16.50 |

Hence from the above data K = 16.50 Newton/ meter =
K(Theoretical)

From the above graph slope comes out to be : 17.085 = K(Observed)

Hence K = 17.08Newton/metre

% error = [K(Theo) -K(Observed)]/K(Observed) * 100

= [16.50-17.08]/17.08 *100

=3.39 %

{Note : We have neglected -negative sign}

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