A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
Practice Questions
1 question
Q1
A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
0.5 s
1 s
2 s
4 s
The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.
Questions & Step-by-step Solutions
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Q
Q: A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
Solution: The period T is given by T = 2π√(m/k). Here, T = 2π√(2/200) = 2π√(0.01) = 2π(0.1) = 0.2π ≈ 0.63 s.
Steps: 7
Step 1: Identify the formula for the period of oscillation T in simple harmonic motion, which is T = 2π√(m/k).
Step 2: Identify the values given in the problem: the mass m = 2 kg and the spring constant k = 200 N/m.
Step 3: Substitute the values into the formula: T = 2π√(2/200).
Step 4: Simplify the fraction inside the square root: 2/200 = 0.01.
Step 5: Take the square root of 0.01: √(0.01) = 0.1.
Step 6: Multiply by 2π: T = 2π(0.1).
Step 7: Calculate the final result: T = 0.2π, which is approximately 0.63 seconds.
