A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg

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Question: A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?

Options:

Correct Answer: 1 s

Solution:

T = 2π√(m/k) = 2π√(2/200) = 1 s.

A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?

A simple harmonic oscillator has a spring constant of 200 N/m and a mass of 2 kg. What is its period?

Step 1: Identify the formula for the period (T) of a simple harmonic oscillator, which is T = 2π√(m/k).
Step 2: Identify the values given in the problem: the mass (m) is 2 kg and the spring constant (k) is 200 N/m.
Step 3: Substitute the values into the formula: T = 2π√(2/200).
Step 4: Calculate the fraction inside the square root: 2/200 = 0.01.
Step 5: Take the square root of 0.01, which is 0.1.
Step 6: Multiply by 2π: T = 2π * 0.1.
Step 7: Calculate 2π * 0.1, which is approximately 0.6283.
Step 8: Round the result to the nearest whole number, which gives T ≈ 1 s.

Simple Harmonic Motion – The behavior of oscillating systems where the restoring force is proportional to the displacement.
Period of Oscillation – The time taken for one complete cycle of motion in a simple harmonic oscillator, calculated using the formula T = 2π√(m/k).