Final answer:
Doubling the mass of a simple harmonic oscillator results in an increase of the time period by a factor of √2, indicating that the oscillations will be slower.
Step-by-step explanation:
If the mass attached to a spring becomes double, the effect on its time period can be explained by the formula for the period T of a simple harmonic oscillator: T = 2π√(m/k), where m is the mass and k is the spring constant. By doubling the mass, the new period T' becomes T' = 2π√(2m/k), which simplifies to T' = √2 × T. This shows that the period increases by a factor of √2, which is approximately 1.414, meaning the oscillation will be slower.