A 50 -g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is \(15 \mathrm{m} / \mathrm{s}^{2}\) and its maximum speed is \(3.5 \mathrm{m} / \mathrm{s}\). Determine the (a) angular frequency, (b) spring constant, and (c) amplitude.

Using the formula for the maximum speed in SHM, \(v_{max} = \omega * A\), where \(v_{max}\) is the maximum speed, \(\omega\) is the angular frequency and \(A\) is the amplitude, we can isolate \(\omega\) to get \(\omega = \frac{v_{max}}{A}\). We don't know \(A\) at this point, but we also know that the maximum acceleration in SHM is given by \(a_{max} = \omega^2 * A\), where \(a_{max}\) is the maximum acceleration. Again, we can isolate \(\omega\) to get \(\omega = \sqrt{\frac{a_{max}}{A}}\). Since the two expression should be equal, we equate and solve for \(A\) : \( \frac{v_{max}}{A} = \sqrt{\frac{a_{max}}{A}}\). Squaring both sides and simplifying, we get \( A = \frac{v_{max}^2}{a_{max}}\). Substituting the provided values gives \(A = \frac{(3.5m/s)^2}{15m/s^2} = 0.82 m\). Now we can substitute \(A = 0.82 m\) into the first formula to find \(\omega\): \(\omega = \frac{3.5m/s}{0.82m} = 4.27 rad/s\)

The spring constant (\(k\)) can be determined using the formula for angular frequency in SHM, which is \(\omega = \sqrt{\frac{k}{m}}\), where \(m\) is the mass. Rearrange this formula to solve for \(k\): \(k = m * \omega^2\). Substituting the provided mass and the calculated angular frequency gives \(k = 0.05kg * (4.27 rad/s)^2 = 0.91 N/m\)

The amplitude was already calculated when determining the angular frequency. Thus the amplitude of the object's motion is 0.82 m.