A mass attached to a spring oscillates with a period of 3.04 s. If the mass starts from rest at x = 0.0420 m and time t = 0, where is it at time t = 2.75 s?

To find the position of the mass at a given time, we can use the equation for simple harmonic motion:

x(t) = A * cos(ωt + φ)

Where:
- x(t) is the position of the mass at time t.
- A is the amplitude of the motion.
- ω is the angular frequency.
- φ is the phase constant.

We can find the values of A and ω using the given information. The period of the oscillation is given as 3.04 s. The period is related to the angular frequency as:

T = 2π/ω

Solving for ω, we have:

ω = 2π/T = 2π/3.04 ≈ 2.07 rad/s

The amplitude A is given as 0.0420 m.

Now, to find the phase constant φ, we need additional information about the initial conditions. The problem states that the mass starts from rest at x = 0.0420 m and time t = 0. This implies that the initial phase is 0, as the mass starts at the equilibrium position.

With these values, we can now find the position of the mass at time t = 2.75 s:

x(2.75) = 0.0420 * cos(2.07 * 2.75 + 0)

Evaluating this expression, we find:

x(2.75) ≈ 0.0420 * cos(5.6925) ≈ 0.0420 * (-0.937) ≈ -0.0393 m

Therefore, the mass is approximately located at x = -0.0393 m at time t = 2.75 s.