A block with mass 0.55 kg on a frictionless surface is attached to a spring with spring constant 43 N/m. The block is pulled from the equilibrium position and released. What is the period of the system?
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To find the period of the system, we can use the formula for the period of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the mass of the block, and k is the spring constant.
Given:
Mass of the block, m = 0.55 kg
Spring constant, k = 43 N/m
Plugging in the values into the formula, we get:
T = 2π√(0.55 kg / 43 N/m)
Now, let's calculate the result step by step:
Step 1: Divide the mass of the block by the spring constant
0.55 kg / 43 N/m = 0.0128 kg/m
Step 2: Take the square root of the result
√(0.0128 kg/m) = 0.1132 s
Step 3: Multiply the result by 2π (approximately 6.28)
0.1132 s * 6.28 = 0.7104 s
Therefore, the period of the system is approximately 0.7104 seconds.