Final answer:
For a mass-spring system undergoing simple harmonic motion with a given mass of 10 g and spring constant of 10 N/m, the next time the mass will be at maximum displacement is approximately 0.3 seconds. Option d is the answer.
Step-by-step explanation:
The question concerns the next time the mass will be at maximum displacement when dealing with simple harmonic motion (SHM) of a mass attached to a spring. The period of oscillation 'T' for a mass-spring system in SHM can be calculated using the formula:
T = 2π √(m/k)
where:
- m is the mass in kilograms (kg),
- k is the spring constant in newtons per meter (N/m).
Given that the mass is 10 g (0.01 kg) and the spring constant is 10 N/m, the period 'T' is:
T = 2π √(0.01 kg / 10 N/m)
T = 2π √(0.001)
T = 2π √(1/1000)
T = 2π / 10
T = 0.2π
Approximating π to 3.14:
T = 0.2 × 3.14 ≈ 0.628 seconds
The mass will thus be at maximum displacement again after half a period, as that is the time it takes to go from one extreme to the other extreme in SHM:
Next maximum displacement = T/2
Next maximum displacement = 0.628 / 2 ≈ 0.314 seconds
The closest answer in the provided options, accounting for rounding, would be:
d. 0.3 seconds