"

Set 9 Problem number 10


Problem

An ideal spring has restoring force constant 160 Newtons/meter. An unknown mass on a spring is observed to complete a cycle of oscillation in 1.77 seconds. What is the mass, in kilograms?

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 1.77 seconds. Thus the object completes 2 `pi radians in 1.77 seconds. This implies a rate of

angular frequency = 2 `pi radians/ 1.77 seconds = 3.549 radians/second.

Since we know k, we know that 3.549 radians/second = `sqrt[( 160 Newtons/meter) / m].

Solving for m we obtain m = ( 160 Newtons/meter) / ( 3.549 radians/second) ^ 2 = 12.7 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 Solution

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.

Explanation in terms of Figure(s), Extension

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.

Figure(s)

k_omega_and_m.gif (4415 bytes)

"