Final answer:
The period of a simple harmonic oscillator will change by a factor of approximately 1/1.41 (or 1/√2) if the spring constant is doubled, not by any of the provided factors of 2, 1/2, 4, or 1/4.
Step-by-step explanation:
When assessing how the spring change affects the period of a mass on a spring, we know from physics and the formula for the period of a simple harmonic oscillator (T = 2π√(m/k)) that the period, T, is directly proportional to the square root of the mass, m, and inversely proportional to the square root of the spring constant, k. If the spring constant is doubled, the new period, T', would be T' = 2π√(m/(2k)) = 2π√(m/k)/√2 = T/√2. Therefore, the period would change by a factor of 1/√2, which is approximately 1/1.41, not by a factor of 2, 1/2, 4, or 1/4. Since neither of the provided options 1) The period would change by a factor of 2, 2) The period would change by a factor of 1/2, 3) The period would change by a factor of 4, nor 4) The period would change by a factor of 1/4 is correct, there appears to be a discrepancy in the options given.