a mass of 0.2kg is attached to a spring of negligible mass. if the mass executes simple harmonic motion with a period of 0.5s, what will be the spring constant?
To find the spring constant, we can use the formula for the period of simple harmonic motion:
T = 2π√(m/k)
Where:
T = period of oscillation
m = mass attached to the spring
k = spring constant
Given that the mass (m) is 0.2 kg and the period (T) is 0.5 s, we can rearrange the formula to solve for the spring constant (k).
Rearranging the formula:
T = 2π√(m/k)
(T/2π)² = m/k
k = m/(T/2π)²
Plugging in the values:
k = 0.2 / ((0.5 / (2π))²)
Calculating the spring constant:
k = 0.2 / ((0.5 / 6.28)²)
k ≈ 1.005 N/m
Therefore, the spring constant (k) is approximately 1.005 N/m.
To calculate the spring constant, we can use the formula of the period of simple harmonic motion which involves the mass and spring constant.
Given:
Mass (m) = 0.2 kg
Period (T) = 0.5 s
The period (T) of simple harmonic motion is related to the spring constant (k) and mass (m) by the formula:
T = 2π √(m/k)
Rearranging the formula, we can solve for the spring constant (k):
k = (4π² m) / T²
Now, plug in the given values into the formula:
k = (4π² * 0.2 kg) / (0.5 s)²
Calculating:
k = (4π² * 0.2 kg) / (0.25 s²)
k = (4π² * 0.2 kg) / 0.0625 s²
k = (4 * 3.14159² * 0.2 kg) / 0.0625 s²
k ≈ 161.03 N/m (spring constant)
Therefore, the spring constant for this system is approximately 161.03 N/m.