What is the mass in kilograms of an object if, when suspended from an ideal spring whose restoring force constant is 560 Newtons/meter, the mass completes a cycle of oscillation in 1.62 seconds.
We know that the angular frequency of an object in simple harmonic motion is `sqrt(k/m). The angular frequency is in radians/second.
The information given is that the object completes a cycle in 1.62 seconds. Thus the object completes 2 `pi radians in 1.62 seconds. This implies a rate of
angular frequency = 2 `pi radians/ 1.62 seconds = 3.878 radians/second.
Since we know k, we know that 3.878 radians/second = `sqrt[( 560 Newtons/meter) / m].
Solving for m we obtain m = ( 560 Newtons/meter) / ( 3.878 radians/second) ^ 2 = 37.23 kilograms.
In symbols, we solve `omega = `sqrt(k/m) for m, obtaining m = k / `omega^2, then substitute the known value of k and the value of `omega found above.
If the time for a oscillation is T, then the reference point goes around the circle in time T. Since the circle corresponds to 2 `pi radians, this motion corresponds to an angular frequency of
angular frequency = `omega = 2 `pi / T.
Using the fundamental relationship `omega = `sqrt( k / m), we solve for m to obtain
mass of oscillating object = m = k / `omega^2.
With the value we obtained for `omega, this allows us to determine m.
The figure shows the relationship between `omega, k and m (the blue triangle) and the relationships among `omega, T and f. Knowing T we find `omega; knowing `omega and k we then find m.
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