A .255 kg mass is attached to a spring and executes simple harmonic motion with a period of .56 s. The total energy of the system is 3.6 J.
Find the force constant of the spring.
Answer in units of N/m
To find the force constant of the spring, we can use the formula for the period of simple harmonic motion with a mass-spring system:
T = 2π √(m/k)
where T is the period, m is the mass, and k is the force constant of the spring.
We are given the period T = 0.56 s and the mass m = 0.255 kg. We can rearrange the formula to solve for k:
k = (4π² m) / T²
Substituting the given values:
k = (4π² * 0.255 kg) / (0.56 s)²
Now, let's calculate it step by step:
Step 1: Calculate the denominator: (0.56 s)²
= 0.3136 s²
Step 2: Calculate the numerator: (4π² * 0.255 kg)
= 10.146π² kg
Step 3: Divide the numerator by the denominator:
k = (10.146π² kg) / (0.3136 s²)
Step 4: Approximate the value of π (pi) using 3.14159:
k ≈ (10.146 * 3.14159² kg) / (0.3136 s²)
Step 5: Solve for k:
k ≈ (10.146 * 9.8696 kg) / (0.3136 s²)
Step 6: Calculate the numerical value:
k ≈ 100.102 N/m
Therefore, the force constant of the spring is approximately 100.102 N/m.