When a mass of 23 g is attached to a certain
spring, it makes 16 complete vibrations in
4.6 s.
What is the spring constant of the spring?
Answer in units of N/m.
To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement of the mass from its equilibrium position.
Hooke's Law can be written as:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement.
In this case, we know that the mass attached to the spring makes 16 complete vibrations in 4.6 seconds.
First, let's calculate the time period of one complete vibration:
Time period (T) = Total time / Number of vibrations
T = 4.6 s / 16 = 0.2875 s
Next, let's find the angular frequency (ω) of the vibrations:
ω = 2π / T
ω = 2π / 0.2875 s = 21.863 rad/s
Now, let's calculate the mass in kilograms:
mass (m) = 23 g = 0.023 kg
Using the formula for the angular frequency of a mass-spring system:
ω = √(k / m)
we can rearrange it to solve for the spring constant:
k = m * ω^2
k = 0.023 kg * (21.863 rad/s)^2
k ≈ 10.14 N/m
Therefore, the spring constant of the spring is approximately 10.14 N/m.