Question: A 2.0 kg mass attached to a spring oscillates back and forth along a frictionless surface when a force of 12 N is applied. The mass reaches a maximum velocity of 0.25 m/s and the maximum displacement from equilibrium is 0.025 m. Calculate the amplitude, frequency, period, and maximum acceleration of the wave.

Answer: Given: Mass (m) = 2.0 kg Force (F) = 12 N Maximum velocity (v) = 0.25 m/s Maximum displacement (x) = 0.025 m

Firstly, let's find the amplitude (A). The maximum displacement is the amplitude, so A = 0.025 m.

Next, let's calculate the frequency (f). Frequency can be calculated using the formula: f = v / λ where v = velocity and λ = wavelength In this case, the wavelength is the distance covered by the mass in one complete cycle. Since the mass moves from equilibrium to maximum displacement in one complete cycle: λ = 2 * A λ = 2 * 0.025 λ = 0.05 m

We can now calculate the frequency: f = v / λ f = 0.25 / 0.05 f = 5 Hz

Moving on to find the period (T), The period can be found using the formula: T = 1 / f T = 1 / 5 T = 0.2 s

Finally, let's determine the maximum acceleration (a). Maximum acceleration occurs at maximum displacement and can be found using the formula: a = (2πf)^2 * A

Substitute the known values: a = (2 * π * 5)^2 * 0.025 a = (10π)^2 * 0.025 a ≈ 785.4 m/s^2

Therefore, the amplitude is 0.025 m, frequency is 5 Hz, period is 0.2 s, and maximum acceleration is approximately 785.4 m/s^2.