A block has mass 800 g, whose is fasten to a spring with spring constant k is 70 N/m.
The block is pulled a distance x = 15 cm from its equilibrium position at x = 0 on a
frictionless surface and released from rest at t = 0.
(a) What force does the spring exert on the block just before the block is released?
(b) What are the angular frequency, the frequency, and the period of the resulting
oscillation?
(c) What is the amplitude of the oscillating block?
(d) What is the maximum speed of the oscillating block?
(e) What is the magnitude of the maximum acceleration of the block?
Start out by reviewing Hooke's Law.
The frequency of motion will be
f = 2*pi*sqrt(k/m) Hz
The amplitude A will be the initial deflection.
The maximum velocity is w * A
w = sqrt(k/m)
You do the rest. You should not be deprived of the learning experience.
Thanks i will do the rest
I got the frequency equation wrong.
It is f = [1/(2 pi)] sqrt (k/m)
The constant factor was upside down
To answer these questions, we'll need to use the equations of motion for a mass-spring system. Let's go step by step:
(a) What force does the spring exert on the block just before the block is released?
The force exerted by the spring can be calculated using Hooke's Law, which states that the force is proportional to the displacement from the equilibrium position. The equation for this force is:
F = -kx
where F is the force, k is the spring constant, and x is the displacement.
In this case, the block is pulled 15 cm from the equilibrium position, which corresponds to a displacement of x = 0.15 m. Substituting this value into the equation, we get:
F = -(70 N/m)(0.15 m) = -10.5 N
Note that the negative sign indicates that the force is in the opposite direction of the displacement.
(b) What are the angular frequency, the frequency, and the period of the resulting oscillation?
To find the angular frequency (ω), we can use the formula:
ω = √(k / m)
where ω is the angular frequency, k is the spring constant, and m is the mass of the block.
In this case, the mass of the block is given as 800 g, which is equal to 0.8 kg. Substituting the values into the equation, we get:
ω = √(70 N/m / 0.8 kg) = √87.5 rad/s ≈ 9.354 rad/s
To find the frequency (f), we can use the formula:
f = ω / (2π)
Substituting the value of ω, we get:
f ≈ 9.354 rad/s / (2π) ≈ 1.49 Hz
The period (T) of the oscillation is the reciprocal of the frequency, so:
T = 1 / f ≈ 1 / 1.49 Hz ≈ 0.671 s
(c) What is the amplitude of the oscillating block?
The amplitude (A) of the oscillation can be determined from the initial displacement. In this case, the block is pulled 15 cm from the equilibrium position, so the amplitude is:
A = 0.15 m
(d) What is the maximum speed of the oscillating block?
The maximum speed of the oscillating block can be found using the formula:
v_max = Aω
where v_max is the maximum speed and A is the amplitude.
Substituting the values, we get:
v_max = (0.15 m) * (9.354 rad/s) ≈ 1.403 m/s
(e) What is the magnitude of the maximum acceleration of the block?
The maximum acceleration (a_max) of the block can be found using the formula:
a_max = Aω^2
where a_max is the maximum acceleration and A is the amplitude.
Substituting the values, we get:
a_max = (0.15 m) * (9.354 rad/s)^2 ≈ 20.742 m/s^2