Problem: What is the spring constant of a spring which oscillates at 2.9 cycles/second when a mass of 2 kilograms is placed on it?
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Solution: We know that the angular frequency of an object in simple harmonic motion is `sqrt(k/m).
The information given is that the object completes 2.9 cycles every second. Thus the object completes 2 `pi ( 2.9) radians each second. This implies a rate of 2 `pi ( 2.9) radians/second = 18.22 radians/second. This is the angular frequency.
Since we know m, we know that 18.22 radians/second = `sqrt[k/( 2 kg)].
Solving for k we obtain k = 2 kg( 18.22 radians/second) ^ 2 = 663.9 N/m.
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.