Since it undergoes simple harmonic motion, its displacement versus time is a sine or cosine function. Since we are given that at \( t = 0 \, \text{s} \), the displacement is at a minimum, we know it will be a negative cosine curve. Looking at this graph, we can see that after \( \frac{3}{4} \, T \), it will be at \( 0 \), hence the answer is \( B \).
CloseThe acceleration of an object is given by \( a = \omega^2 \cdot x \), where \( \omega = \frac{2\pi}{T} \). Substituting \( \omega \) into the equation gives \( a = \left(\frac{2\pi}{T}\right)^2 \cdot x_0 = \frac{4\pi^2}{T^2}x_0 \). Hence, the answer is \( D \).
CloseThey only point in the same direction when the object moves from the extreme position to the mean position, which happens twice in 1 period. The other 2 times, it moves from the mean to the extreme position, during which they are opposite. Hence, the answer is \( B \).
CloseThe acceleration is given by \( a = -kx \), which shows that when the displacement is the largest, the acceleration is the largest. This occurs only at the two extremes. Hence, the answer is \( C \).
CloseLogically thinking, it cannot have the largest velocity at the two extremes since that is where it stops momentarily and changes directions. By method of elimination, it must be the equilibrium position. Additionally, the restoring force always points towards the equilibrium position, and the object speeds up the most as it moves toward this middle point. Once it passes the equilibrium, it begins to slow down as it moves toward the extremes. Hence, the answer is \( B \).
CloseThe period of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \). Substituting \( 2m \) for \( m \) gives:
\[ T_{\text{new}} = 2\pi \sqrt{\frac{2m}{k}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{m}{k}} = \sqrt{2}T \]
CloseThe total energy of the system cannot change, since it is assumed to be closed (no losses to friction etc.), so it must be A. As the object moves, energy is transferred from potential to kinetic, and vice versa, but their sum never changes.
CloseThe period of oscillation for a mass-spring system is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Substituting in the values: \[ T = 2\pi \sqrt{\frac{0.2}{50}} \] \[ T = 2\pi \sqrt{0.004} \approx 2\pi \times 0.063 \approx 0.40 \, \text{s} \]
Close\[ v_{\text{max}} = \omega A \] \[ v_{\text{max}} = 5.0 \times 0.1 = 0.5 \, \mathrm{m/s} \]
Close\[ E = \frac{1}{2} k A^2 \] \[ E = \frac{1}{2} \times 80 \times (0.2)^2 = \frac{1}{2} \times 80 \times 0.04 = 1.6 \, \mathrm{J} \]
Close\[ a_{\text{max}} = \omega^2 A \] \[ a_{\text{max}} = (2.0)^2 \times 0.3 = 4.0 \times 0.3 = 1.2 \, \mathrm{m/s^2} \]
Close\[ x(t) = A \cos(\omega t) \] \[ x(2) = 0.5 \cos(3.0 \times 2) = 0.5 \cos(6.0) \] \[ \cos(6.0 \, \mathrm{rad}) \approx 1 \quad \Rightarrow x(2) \approx 0.5 \, \mathrm{m} \]
CloseThe maximum kinetic energy equals the total energy, which is the potential energy at maximum displacement: \[ E = \frac{1}{2} k A^2 \] \[ E = \frac{1}{2} \times 100 \times (0.4)^2 = \frac{1}{2} \times 100 \times 0.16 = 8.0 \, \mathrm{J} \]
CloseThe period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{L}{g}} \] \[ T = 2\pi \sqrt{\frac{2.0}{9.8}} = 2\pi \sqrt{0.204} \approx 2\pi \times 0.452 = 2.84 \, \mathrm{s} \]
CloseIn simple harmonic motion: