A 4.5-kg block is hung from a spring causing the spring to elongate 12 cm. With what period will the spring oscillate if the spring was stretched 15 cm and released?
for your other mass/spring problem I derived:
(2 pi f)^2 = k/m
same here
to get k:
F = weight = 4.5 * 9.81 = 44.1 N
deflection = .12 m
so k = 44.1/.12
now solve for f = (1/2pi)sqrt(k/m)
and
T = 1/f
To calculate the period of oscillation, we need to use Hooke's Law and Newton's second law of motion. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Newton's second law states that the force acting on an object is equal to its mass multiplied by its acceleration.
First, let's find the spring constant, k:
1. We know that the spring elongates 12 cm (0.12 m) when a 4.5 kg block is hung from it. We can use Hooke's Law to calculate the spring constant:
F = k * x
where F is the force, x is the displacement, and k is the spring constant.
Rearranging the equation, we have:
k = F / x
The force can be calculated using Newton's second law:
F = m * g
where m is the mass of the block (4.5 kg), and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we have:
F = 4.5 kg * 9.8 m/s^2 = 44.1 N
Thus, the spring constant is:
k = 44.1 N / 0.12 m ≈ 367.5 N/m
Next, let's calculate the angular frequency, ω:
2. The angular frequency, ω, is related to the mass, m, and the spring constant, k, through the equation:
ω = √(k / m)
Substituting the values, we have:
ω = √(367.5 N/m / 4.5 kg) ≈ 9.44 rad/s
Finally, let's find the period, T, of oscillation:
3. The period, T, is the reciprocal of the angular frequency:
T = 2π / ω
Substituting the value of ω, we have:
T = 2π / 9.44 rad/s ≈ 0.668 s
Therefore, the spring will oscillate with a period of approximately 0.668 seconds when stretched 15 cm and released.