# mass spring system

mass spring system experiment

plot 左個graph T^2 against m

### 1 個解答

• 天同
Lv 7
1 十年前
最愛解答

This is mainly because the "effective mass" of the spring has not been taken into account.

When the mass oscillates, part of the spring also performs simple harmonic motion. It has been shown that the part of spring that contributes to oscillation equals to one-third of the spring mass.

Thus, the equation that you plot, T^2 = (2.pi)^2.[m/k] should be re-wriiten as

T^2 = (2.pi)^2[(m+M/3)/k]

where T is the [eriod of oscillation

pi = 3.14159.....

m is the mass of the weight under oscillation

M is the mass of spring

k is the spring constant

Hence, T^2 = 4.(pi)^2(m/k) + 4.(pi)^2.[M/3k]

Since 4.(pi)^2.[M/3k] is a constant, this explains why the graph of T^2 against m is not passing through the origin.

But there should be a +ve y-intercept instead of a +ve x-intercept (which gives a -ve y-intercept) as you found.

The reson for this, apart from experimental uncertainty, is because of the reason that the spring doesn't strictly obey Hooke's Law, especially at small extension. This situation usually would happen in aged spring. The spring is not able to restore to its original length after extension. Therefore, although the period is zero (T= 0 s, no oscillation), the x-intercept shows a finite mass. That is, the spring is still elongated as if there is some mass hanging to it.