Problem: A mass on a spring is observed to complete a cycle of oscillation in 2.4 seconds. The spring constant is 23 Newtons/meter. What is the mass, in kilograms?
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Solution: 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 2.4 seconds. Thus the object completes 2 `pi radians in 2.4 seconds. This implies a rate of
angular frequency = 2 `pi radians/ 2.4 seconds = 2.617 radians/second.
Since we know k, we know that 2.617 radians/second = `sqrt[( 23 Newtons/meter) / m].
Solving for m we obtain m = ( 23 Newtons/meter/)( 2.617 radians/second) ^ 2 = 3.358 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.
Generalized Response: 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.
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Figure Description: 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.
