Q4 A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65N/m The block is pulled a distance x(=11cm) from its equilibrium position and released from rest. (a) What force does the spring exert on the block just before it is released? (3 marks) (b)What are the angular frequency , the frequency, and the period of oscillation? (6 Marks) (c) What is the amplitude of the oscillation? (2 Marks) (d) What is the maximum speed of the oscillating block?(3 marks) (e) What is the magnitude of the maximum acceleration of the block? (3 marks) (3 marks) (f) What is the phase constant of the motion?

(a) The force exerted by the spring on the block just before it is released can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. The formula for Hooke's Law is:<br /><br />F = kx<br /><br />where F is the force, k is the spring constant, and x is the displacement.<br /><br />In this case, the mass of the block is 680 g, which is equal to 0.68 kg. The spring constant is 65 N/m, and the displacement is 11 cm, which is equal to 0.11 m.<br /><br />Substituting these values into the formula, we get:<br /><br />F = (65 N/m)(0.11 m) = 7.15 N<br /><br />Therefore, the force exerted by the spring on the block just before it is released is 7.15 N.<br /><br />(b) The angular frequency, frequency, and period of oscillation can be calculated using the following formulas:<br /><br />Angular frequency (ω) = √(k/m)<br />Frequency (f) = ω/2π<br />Period (T) = 1/f<br /><br />Substituting the given values, we get:<br /><br />ω = √((65 N/m)/(0.68 kg)) = 9.6 rad/s<br />f = ω/2π = 1.53 Hz<br />T = 1/f = 0.65 s<br /><br />Therefore, the angular frequency is 9.6 rad/s, the frequency is 1.53 Hz, and the period is 0.65 s.<br /><br />(c) The amplitude of the oscillation is the maximum displacement of the block from its equilibrium position. In this case, the block is pulled a distance of 11 cm from its equilibrium position and released from rest. Therefore, the amplitude of the oscillation is 11 cm.<br /><br />(d) The maximum speed of the oscillating block can be calculated using the formula:<br /><br />v_max = ωA<br /><br />where v_max is the maximum speed, ω is the angular frequency, and A is the amplitude.<br /><br />Substituting the given values, we get:<br /><br />v_max = (9.6 rad/s)(0.11 m) = 1.06 m/s<br /><br />Therefore, the maximum speed of the oscillating block is 1.06 m/s.<br /><br />(e) The magnitude of the maximum acceleration of the block can be calculated using the formula:<br /><br />a_max = ω^2A<br /><br />where a_max is the maximum acceleration, ω is the angular frequency, and A is the amplitude.<br /><br />Substituting the given values, we get:<br /><br />a_max = (9.6 rad/s)^2(0.11 m) = 10.3 m/s^2<br /><br />Therefore, the magnitude of the maximum acceleration of the block is 10.3 m/s^2.<br /><br />(f) The phase constant of the motion can be calculated using the formula:<br /><br />φ = ωt + θ<br /><br />where φ is the phase constant, ω is the angular frequency, t is the time, and θ is the phase angle.<br /><br />In this case, the block is released from rest, which means that the phase angle is 0. Therefore, the phase constant of the motion is:<br /><br />φ = (9.6 rad/s)(0 s) + 0 = 0<br /><br />Therefore, the phase constant of the motion is 0.