Set 9 Problem number 14

• Problem

• Solution

Problem

What is the restoring force constant of a spring when a mass of .9941 kilograms is placed on it, requires .21 seconds to complete an osciallation?

Solution

We know that the angular frequency of an object in simple harmonic motion is `omega = `sqrt(k/m).

The information given is that the object completes a cycle in .21 seconds.

• Thus the object completes 2 `pi ( 4.761) radians in .21 seconds.
• This implies a rate of

• angular frequency = 2 `pi radians/ ( .21 seconds) = 29.92 radians/second.

Since we know m, we know that 29.92 radians/second = `sqrt[k/( 2 kg)].

• Solving for k we obtain k = .9941 kg( 29.92 radians/second) ^ 2 = 890 N/m.

The symbolic solution tell us that if `omega = `sqrt(k / m), then `omega^2 = k / m and m = k / `omega^2.

• Substituting the known k and `omega we obtain the same result as before.

In symbols, we solve `omega = `sqrt(k/m) for k, obtaining k = m * `omega ^ 2, then substitute the known values of m and the `omega found above.