A spring is hung from the ceiling. A 0.300 kg block is then attached to the free end of the spring. When released from rest, the block drops 0.170 m before momentarily coming to rest.
(a) What is the spring constant of the spring?
(b) Find the angular frequency of the block's vibrations.
To find the spring constant of the spring and the angular frequency of the block's vibrations, we can use the equations of motion and Hooke's law.
(a) To find the spring constant, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. In equation form, this is F = -kx, where F is the force, k is the spring constant, and x is the displacement.
Given that the block drops 0.170 m before momentarily coming to rest, we can calculate the force acting on it using Newton's second law, F = mg, where m is the mass of the block and g is the acceleration due to gravity.
F = mg
F = 0.300 kg * 9.8 m/s^2
F = 2.94 N
Since the block is momentarily at rest at the maximum displacement, the force exerted by the spring must balance the force of gravity. Therefore, we have:
-kx = mg
Using the values we have:
-k * 0.170 m = 2.94 N
Simplifying further, we find:
k = -2.94 N / 0.170 m
k ≈ 17.29 N/m
So, the spring constant of the spring is approximately 17.29 N/m.
(b) The angular frequency (ω) can be calculated using the equation:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the block.
Using the values we know:
k = 17.29 N/m
m = 0.300 kg
ω = sqrt(17.29 N/m / 0.300 kg)
ω ≈ sqrt(57.63 rad/s^2)
ω ≈ 7.59 rad/s
Therefore, the angular frequency of the block's vibrations is approximately 7.59 rad/s.