With only the simplest equipment, you too can perform mass measurements similar to the ones aboard Skylab.

You will need:

Instructions

The theory predicts that the oscillation period should be proportional to the square root of the oscillating mass, including the mass of the clip. Note that gravity plays no part here: the oscillation period would be the same on the Moon or in zerog. Denoting square root by SQRT, we have

(T_{2}/T_{1}) = SQRT(m_{2}+m_{0})/SQRT(m_{1}+m_{0}) = SQRT[(m_{2}+m_{0})/(m_{1}+m_{0})].

("The ratio of square roots is the square root of the ratio"). Muliplying each side by itself: (T_{2}/T_{1})^{2} = (m_{2}+m_{0})/(m_{1}+m_{0}). If we were in space, measured T_{1} and T_{2}, and knew the masses m_{1} and m_{0}, then we could calculate an unknown mass m_{2}.

Weights: m_{1} = 50 gr, m_{2} = 120 gr, m_{0} = 10 gr. The number of oscillations counted in a 10second period was: with m_{1}, 20 oscillations, with m_{2}, 13.5 oscillations. Then

T_{1} = 10 sec/20 = 0.5 sec T_{2} = 10 sec/13.5 = 0.74074 sec. so that
(T_{2}/T_{1})^{2} = 2.195 should equal (m_{2}+m_{0})/(m_{1}+m_{0}) = 130/60 = 2.167 
This agreement is probably better than such a crude experiment deserves, considering that the mass of the sawblade itself was ignored.

Next Stop: #18 Newton's Second Law
Author and curator: David P. Stern
Last updated 3 April 1999