91Ó°ÊÓ

To find the equilibrium position, we set the force due to the spring equal to the weight of the body: \(F_{spring} = k \cdot y_{eq}\), where \(F_{spring}\) is the force due to the spring, \(k\) is the spring constant, and \(y_{eq}\) is the equilibrium position. Since the body's weight is acting downward, its force can be expressed as: \(F_{body} = -4.0 \text{ N}\). At equilibrium, the spring force is equal to the weight of the body: \(F_{spring} = F_{body}\). Hence: \(k \cdot y_{eq} = -4.0 \text{ N}\). Substitute the value of \(k\): \(250 \text{ N/m} \cdot y_{eq} = -4.0 \text{ N}\). Now, solve for \(y_{eq}\): \(y_{eq} = \dfrac{-4.0 \text{ N}}{250 \text{ N/m}} = \dfrac{-1}{62.5}\).

Since the body is in SHM, the equation of motion can be written using the following general equation: \(y(t) = A \cdot \cos(\omega t + \phi)\), where \(y(t)\) is the displacement at time \(t\), \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. To find the amplitude \(A\), we observe that the body is released from rest at its maximum displacement, which is equal to its equilibrium position \(y_{eq}\): \(A = -y_{eq} = \dfrac{1}{62.5}\). The angular frequency \(\omega\) can be found using the following formula: \(\omega = \sqrt{\dfrac{k}{m}}\). We can find the mass \(m\) of the body using the formula for weight: \(F_{body} = m \cdot g \Longrightarrow m = \dfrac{F_{body}}{g} = \dfrac{-4.0 \text{ N}}{-9.81 \text{ m/s}^2} \approx 0.408 \text{ kg}\). Now, substitute the values of \(k\) and \(m\): \(\omega = \sqrt{\dfrac{250 \text{ N/m}}{0.408 \text{ kg}}} \approx 24.7 \text{ rad/s}\). Since the body is released from its maximum displacement at \(t=0\), the phase angle \(\phi\) is 0: \(\phi = 0\).

Now, we have all the necessary information to write the equation of motion: \(y(t) = \dfrac{1}{62.5} \cdot \cos(24.7t + 0)\). The equation of motion for the body is: \(y(t) = \dfrac{1}{62.5} \cdot \cos(24.7t)\).