A mass of 0.2 kg is attached to a spring of negligible mass. I the mass executes simple harmonic motion with a period of 0.5 s, what will be the spring constant/
Can I use the following formula to solve this problem
k = (4pi square x mass) / T square
Not a chemistry course I took.
Yes, you can use the formula you mentioned to find the spring constant (k) in this problem. The formula you provided is a rearranged version of the equation for the period (T) of a mass-spring system in simple harmonic motion. The correct formula for the period is:
T = 2π√(m/k)
Where:
T = Period of oscillation
m = Mass attached to the spring
k = Spring constant
To find the spring constant, we can rearrange this equation as follows:
k = (4π²m) / T²
Given that the mass (m) is 0.2 kg and the period (T) is 0.5 seconds, we can substitute these values into the formula to solve for k:
k = (4π² * 0.2) / (0.5)²
Simplifying further:
k = (4π² * 0.2) / 0.25
k ≈ 4π² * 0.2 * 4
k ≈ 3.183
So, the spring constant in this system would be approximately 3.183 N/m.