Can someone help me solve this? I've tried it a few times, but I keep getting answers that seem off. Like, for example, I'm getting 2.339 for part A, and it just seems completely off. So if you know the right formulas and can walk me through this, I'd appreciate it a lot!

The oscillatory motion of a horizontal spring mass system is described by the following equation: x(t) = (0.10m) sin(2π t/(0.94s) + π/5). The mass is 1.3kg mass and the spring constant is 65N/kg.

(a) Find the maximum speed (in m/s)

(b) Find the position of the mass at t = 0 s in meters.

(c) Find the speed of the mass at t = 0 s in meters

(d) Find the spring potential energy of the system at t = 0 s in Joules.

(e) Find the mechanical energy of the system relative to the equilibrium position at t = 0 s in Joules.

Use 10 N/kg for g.

To solve these problems, we need to use the given equation and the formulas related to oscillatory motion and energy. Let's go through each part step by step:

(a) To find the maximum speed (v_max) of the system, we need to differentiate the given equation with respect to time (t). The derivative of x(t) will give us the velocity (v) as a function of time:

v(t) = dx(t)/dt

Taking the derivative of the equation x(t) = (0.10m) sin(2π t/(0.94s) + π/5) with respect to t, we get:

v(t) = (0.10m) * (2π/(0.94s)) * cos(2π t/(0.94s) + π/5)

Now, to find the maximum speed, we need to find the absolute maximum value of v(t). This occurs when the cosine term reaches its maximum value of 1. Therefore:

v_max = (0.10m) * (2π/(0.94s)) * 1

Simplifying further:

v_max = (2π * 0.10) / 0.94

Evaluating this expression, we get:

v_max ≈ 2.126 m/s

(b) To find the position of the mass (x) at t = 0 s, we can substitute t = 0 into the given equation:

x(0) = (0.10m) sin(2π * 0/(0.94s) + π/5)

Simplifying further:

x(0) = (0.10m) sin(π/5)

Evaluating this expression, we get:

x(0) ≈ 0.052 m

(c) To find the speed (v) of the mass at t = 0 s, we can substitute t = 0 into the velocity equation we derived earlier:

v(0) = (0.10m) * (2π/(0.94s)) * cos(2π * 0/(0.94s) + π/5)

Simplifying further:

v(0) = (0.10m) * (2π/(0.94s)) * cos(π/5)

Evaluating this expression, we get:

v(0) ≈ 1.948 m/s

(d) To find the spring potential energy (U_spring) of the system at t = 0 s, we can use the formula:

U_spring = (1/2) * k * x^2

where k is the spring constant and x is the displacement from the equilibrium position. At t = 0 s, we already found x(0) ≈ 0.052 m and k = 65 N/kg. Substituting these values into the formula, we get:

U_spring = (1/2) * 65 N/kg * (0.052 m)^2

Calculating this expression, we get:

U_spring ≈ 0.086 J

(e) To find the mechanical energy (E_mechanical) of the system relative to the equilibrium position at t = 0 s, we need to consider the sum of the kinetic energy and potential energy. The equation is given by:

E_mechanical = (1/2) * m * v^2 + U_spring

where m is the mass of the system. At t = 0 s, we already found m = 1.3 kg, v(0) ≈ 1.948 m/s, and U_spring ≈ 0.086 J. Substituting these values into the formula, we get:

E_mechanical = (1/2) * 1.3 kg * (1.948 m/s)^2 + 0.086 J

Calculating this expression, we get:

E_mechanical ≈ 3.047 J

Note: You mentioned to use 10 N/kg for g (acceleration due to gravity), but it is not required to solve these particular problems.