A block of mass m 0.750 kg is fastened to an unstrained hor- izontal spring whose spring constant is k 82.0 N/m. The block is given a displacement of 0.120 m, where the sign indicates that the dis- placement is along the x axis, and then released from rest. (a) What is the force (magnitude and direction) 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.
F = -82 * .12 = -9.84 N, I can not rad your sign so I will assume -
F = m a = -kx
m a = .82 x
let w = 2 pi f = 2 pi /T
if x = A sin w t
dx/dt = v = A w cos w t
a = dv/dt = -A w^2 sin w t = -w^2 x
so m a = - m w^2 x = - k x
w^2 = k/m
w = 2 pi f = sqrt (k/m)
f = ( 1 / 2pi ) sqrt (82 / 0.75) = 1.67 Hz
max v = A w
max a = A w^2
To answer these questions, we need to apply Hooke's Law and the equations of simple harmonic motion.
Step 1: Determine the force exerted by the spring
According to Hooke's Law, the force exerted by a spring is given by F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the displacement is -0.120 m (negative because it is to the left), and the spring constant is 82.0 N/m.
Therefore, the force exerted by the spring is:
F = -kx = -(82.0 N/m)(-0.120 m) = 9.84 N.
The magnitude of the force is 9.84 N, and since the displacement is to the left, the direction of the force is to the right.
Step 2: Find the angular frequency
The angular frequency (ω) of an oscillatory motion is related to the mass (m) of the block and the spring constant (k) by the formula:
ω = √(k / m)
Substituting the given values:
ω = √(82.0 N/m / 0.750 kg) = 12.3 rad/s.
Step 3: Calculate the maximum speed
The maximum speed of the block in simple harmonic motion can be determined using the formula:
v_max = Aω, where A is the amplitude of the motion (the maximum displacement).
In this case, the amplitude is 0.120 m, and ω is 12.3 rad/s.
Therefore, the maximum speed is:
v_max = (0.120 m)(12.3 rad/s) = 1.47 m/s.
Step 4: Determine the maximum acceleration
The maximum acceleration (a_max) of the block in simple harmonic motion can be found using the formula:
a_max = Aω^2, where A is the amplitude and ω is the angular frequency.
Plugging in the values:
a_max = (0.120 m)(12.3 rad/s)^2 = 17.4 m/s^2.
Summary:
(a) The force exerted by the spring just before the block is released is 9.84 N to the right.
(b) The angular frequency of the resulting oscillatory motion is 12.3 rad/s.
(c) The maximum speed of the block is 1.47 m/s.
(d) The magnitude of the maximum acceleration of the block is 17.4 m/s^2.
To answer these questions, we can use the equations of motion for a mass-spring system. Let's go step by step to find the answers.
(a) The force that the spring exerts on the block just before it is released is given by Hooke's Law:
F = -kx
Where F is the force, k is the spring constant, and x is the displacement. In this case, x = -0.120 m (since it's given that the displacement is along the -x axis). Plugging in the values, we get:
F = -(82.0 N/m)(-0.120 m) = 9.84 N
The magnitude of the force is 9.84 N, and since the displacement is along the -x axis, the direction of the force is in the positive x-axis direction.
(b) The angular frequency (ω) of the resulting oscillatory motion can be found using the formula:
ω = √(k/m)
Where k is the spring constant and m is the mass. Plugging in the values, we get:
ω = √(82.0 N/m / 0.750 kg) = 12.05 rad/s
The angular frequency is 12.05 rad/s.
(c) The maximum speed of the block can be found using the formula:
v_max = Aω
Where A is the amplitude of the oscillation (which is equal to the displacement x) and ω is the angular frequency. Plugging in the values, we get:
v_max = (0.120 m)(12.05 rad/s) = 1.44 m/s
The maximum speed of the block is 1.44 m/s.
(d) The maximum acceleration of the block can be found using the formula:
a_max = Aω^2
Where A is the amplitude of the oscillation (displacement x) and ω is the angular frequency. Plugging in the values, we get:
a_max = (0.120 m)(12.05 rad/s)^2 = 17.45 m/s^2
The magnitude of the maximum acceleration of the block is 17.45 m/s^2.
Note: Make sure to use consistent units throughout the calculations.