Let's solve each part step by step: --- ### **Given:** - Mass, \( m = 0.85 \) kg - Spring constant, \( k = 85 \) N/m - Released from rest at \( d = 0.35 \) m below equilibrium - Position: \( y(t) = A \cos(\omega t - \phi) \) - Positive \( y \)-axis is upward - At \( t = 0 \), \( y(0) = -0.35\, \text{m} \) (since it's below equilibrium, negative) - At \( t = 0 \), released from rest: \( v(0) = 0 \) --- ## **Part (a): Find the angular frequency \(\omega\) in rad/s** For a mass-spring system: \[ \omega = \sqrt{\frac{k}{m}} \] Plug in the values: \[ \omega = \sqrt{\frac{85}{0.85}} = \sqrt{100} = \boxed{10\, \text{rad/s}} \] --- ## **Part (b): Find the magnitude of the coefficient \(A\) in meters** The equation is: \[ y(t) = A \cos(\omega t - \phi) \] At \( t = 0 \): \[ y(0) = A \cos(-\phi) = A \cos \phi \] Also, at \( t = 0 \), \( v(0) = 0 \): \[ v(t) = -A \omega \sin(\omega t - \phi) \] \[ v(0) = -A \omega \sin(-\phi) = A \omega \sin \phi = 0 \] So, \( \sin \phi = 0 \implies \phi = 0 \) or \( \pi \) Let's check which value is correct: - If \( \phi = 0 \), \( y(0) = A \). Then \( A = -0.35 \) m (since initial position is below equilibrium). - If \( \phi = \pi \), \( y(0) = -A \). Then \( -A = -0.35 \implies A = 0.35 \) m. Amplitude is always positive, so take \( A = 0.35 \) m. \[ \boxed{A = 0.35\, \text{m}} \] --- ## **Part (c): Find the value of \(\phi\) in radians** From above: If \( y(0) = -0.35\, \text{m} \) and \( A = 0.35\, \text{m} \), \[ y(0) = A \cos(-\phi) = 0.35 \cos(-\phi) = -0.35 \] \[ \cos(-\phi) = -1 \implies -\phi = \pi \implies \phi = -\pi \] But phase is typically given from \( 0 \) to \( 2\pi \), so use \( \phi = \pi \). \[ \boxed{\phi = \pi} \] --- ## **Part (d): Expression for velocity as a function of time** \[ y(t) = A \cos(\omega t - \phi) \] \[ v(t) = \frac{dy}{dt} = -A \omega \sin(\omega t - \phi) \] \[ \boxed{v(t) = -A\omega \sin(\omega t - \phi)} \] --- ## **Part (e): What is the mass's velocity at \( t_1 = 0.25\, \text{s} \)?** Plug in: - \( A = 0.35 \) m - \( \omega = 10 \) rad/s - \( \phi = \pi \) - \( t_1 = 0.25\, \text{s} \) \[ v(0.25) = -0.35 \times 10 \times \sin(10 \times 0.25 - \pi) \] \[ = -3.5 \times \sin(2.5 - \pi) \] Recall: \( \sin(x - \pi) = -\sin x \) \[ \sin(2.5 - \pi) = -\sin(2.5) \] \[ v(0.25) = -3.5 \times [-\sin(2.5)] = 3.5 \sin(2.5) \] Calculate \( \sin(2.5) \): \[ \sin(2.5) \approx 0.5985 \] \[ v(0.25) = 3.5 \times 0.5985 \approx 2.0948\, \text{m/s} \] **Final answer:** \[ \boxed{v(0.25) \approx 2.09\, \text{m/s}} \] --- ## **Part (f): Magnitude of maximum acceleration** Maximum acceleration for simple harmonic motion: \[ a_{\text{max}} = \omega^2 A \] Plug in the values: - \( \omega = 10 \) rad/s - \( A = 0.35 \) m \[ a_{\text{max}} = (10)^2 \times 0.35 = 100 \times 0.35 = 35\, \text{m/s}^2 \] \[ \boxed{a_{\text{max}} = 35\, \text{m/s}^2} \] --- ## **Summary of answers** **(a)** \( \omega = 10\, \text{rad/s} \) **(b)** \( A = 0.35\, \text{m} \) **(c)** \( \phi = \pi\, \text{rad} \) **(d)** \( v(t) = -A \omega \sin(\omega t - \phi) \) **(e)** \( v(0.25\,\text{s}) \approx 2.09\, \text{m/s} \) **(f)** \( a_{\text{max}} = 35\, \text{m/s}^2 \)