Final answer:
To find the time for an object in SHM to move from x = 0.0 cm to x = 7.6 cm, we can solve the equation x(t) = A × cos(ωt), with A being the amplitude and ω the angular frequency, resulting in the fraction of the period taken for this movement.
Step-by-step explanation:
The question pertains to calculating the time taken for an object in simple harmonic motion (SHM) to move from the equilibrium position (x = 0.0 cm) to a displacement of x = 7.6 cm. Given that the period (T) is 4.0 seconds and the amplitude (A) is 20 cm, we can first determine the angular frequency (ω) of the motion, which is ω = 2π/T. Then, we can use the cosine function to describe the motion as x(t) = A × cos(ωt + ϕ), where ϕ is the phase constant.
When the object is at x = 0.0 cm, the cosine function equals 1, and we are at the equilibrium position at t = 0 when the phase shift is zero. To find the time at x = 7.6 cm, we can solve the equation 7.6 = 20 × cos(2πt/4) for t, which will give us the fraction of the period it takes for the object to reach 7.6 cm from the equilibrium position.
Since the cosine function reaches its maximum at t = 0 and decreases afterwards, the time calculated will reflect the movement from the equilibrium position to 7.6 cm while the block is moving towards the positive amplitude. Considering that this motion is symmetric, we only need to determine the first instance time t, knowing that the motion repeats every 4 seconds.