A spring is hung from the ceiling. A 0.536 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.180 m before momentarily coming to rest, after which it moves back upward. (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 in this problem, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position.
(a) To find the spring constant, we need to calculate the force exerted by the spring. The force exerted by the spring can be determined using the equation:
F = kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this problem, the block drops 0.180 m before momentarily coming to rest, so the displacement x is 0.180 m. The force exerted by the spring when the block is at that position is equal to the weight of the block, given by:
F = mg
where m is the mass of the block and g is the acceleration due to gravity.
In this problem, m = 0.536 kg and g = 9.8 m/s^2.
So, the force exerted by the spring is:
F = (0.536 kg) * (9.8 m/s^2) = 5.2448 N
Now, we can find the spring constant k by rearranging the equation:
k = F / x
k = 5.2448 N / 0.180 m = 29.1378 N/m
Therefore, the spring constant of the spring is 29.1378 N/m.
(b) To find the angular frequency of the block's vibrations, 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 problem, we already found the spring constant k to be 29.1378 N/m, and the mass of the block is 0.536 kg.
So, substituting the values into the formula, we have:
ω = √(29.1378 N/m / 0.536 kg) = √(54.296 Cycles/Second)
≈ 7.366 Cycles/Second (rounded to three significant figures)
Therefore, the angular frequency of the block's vibrations is approximately 7.366 Cycles/Second.
To find the spring constant (k) of the spring, we can use Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
(a) Using Hooke's law, we have:
F = -kx
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
Given that the mass of the block (m) is 0.536 kg and the displacement (x) is 0.180 m, we can consider the force exerted by the spring equal to the weight of the block:
F = mg
Where g is the acceleration due to gravity (9.8 m/s^2).
Substituting the values, we have:
mg = -kx
(0.536 kg)(9.8 m/s^2) = -k(0.180 m)
Now, solve for the spring constant (k):
k = -(0.536 kg)(9.8 m/s^2) / (0.180 m)
k = -5.2488 N/m
The spring constant of the spring is approximately 5.2488 N/m.
(b) The angular frequency (ω) of the block's vibrations can be found using the formula:
ω = √(k/m)
Given that the spring constant (k) is 5.2488 N/m and the mass of the block (m) is 0.536 kg, substitute these values in the formula to find the angular frequency:
ω = √(5.2488 N/m / 0.536 kg)
ω = √(9.78 rad/s^2)
The angular frequency of the block's vibrations is approximately 9.78 rad/s.