A block of mass m = 0.510 kg is fastened to an unstrained horizontal spring whose spring constant is k = 85.3 N/m. The block is given a displacement of +0.112 m, where the + sign indicates that the displacement is along the +x axis, and then released from rest. (a) What is the force (with sign) that the spring exerts on the block just before the block is released? (b) Find the angular frequency of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.
To solve this problem, we can use Hooke's law and the equations of simple harmonic motion.
(a) To find the force that the spring exerts on the block just before it is released, we can use Hooke's law:
F = -kx
where F is the force, k is the spring constant, and x is the displacement. Since the displacement is +0.112 m, the force would be:
F = -(85.3 N/m)(0.112 m) = -9.54 N
The negative sign indicates that the force is in the opposite direction of the displacement.
(b) To find the angular frequency of the resulting oscillatory motion, we can use the equation:
ω = √(k/m)
where ω is the angular frequency, k is the spring constant, and m is the mass. Substituting the given values, we have:
ω = √(85.3 N/m / 0.510 kg) ≈ 6.48 rad/s
(c) To find the maximum speed of the block, we can use the equation for simple harmonic motion:
v_max = Aω
where v_max is the maximum speed, A is the amplitude (which is equal to the displacement in this case), and ω is the angular frequency. Substituting the given values:
v_max = (0.112 m)(6.48 rad/s) ≈ 0.724 m/s
(d) To determine the magnitude of the maximum acceleration of the block, we can use the equation:
a_max = Aω²
where a_max is the maximum acceleration, A is the amplitude (which is equal to the displacement in this case), and ω is the angular frequency. Substituting the given values:
a_max = (0.112 m)(6.48 rad/s)² ≈ 4.87 m/s²