When a 1.3kg mass was attached to a spring, the spring stretched 21cm from its equilibrium position. The mass spring system was then set into oscillatory motion by providing the mass with an initial velocity. Its oscillatory motion is then described by the following equation: Y(t) = (0.09m) sin(2π t/T + π/3).

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

(b) maximum acceleration of the mass (in m/s2),

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

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

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

Part e is really hard for me to find.

The mechanical energy of the system is the sum of kinetic and potential energy. In this ideal world it does not change with time. Since you found the maximum speed use (1/2) m v^2

To find the mechanical energy of the system relative to the equilibrium position at t = 0 s, we need to consider both the potential energy and the kinetic energy of the system.

At any given time, the total mechanical energy E of the system can be calculated as the sum of its potential energy (U) and kinetic energy (K):

E = U + K

Let's break down the calculation step by step:

Step 1: Calculate the potential energy (U)
The potential energy of a mass-spring system is given by the equation:

U = (1/2)kx^2

where k is the spring constant and x is the displacement from the equilibrium position.

In this case, the spring stretches 21 cm from its equilibrium position. Converting this to meters, we have:

x = 0.21 m

We also need to find the spring constant. The spring constant (k) can be calculated using Hooke's Law:

k = F/x

where F is the force exerted by the spring.

From Newton's second law, we know that F = ma, where a is the acceleration of the mass and m is its mass.

Step 2: Calculate the acceleration (a) and spring constant (k)
To find the acceleration, we can differentiate the position equation to obtain the expression for acceleration:

a(t) = d^2Y(t)/dt^2

Taking the second derivative of the position equation gives us:

a(t) = -ω^2Y(t)

where ω is the angular frequency.

Comparing this equation with the standard form of simple harmonic motion (a = -ω^2x), we can see that ω^2 equals the square of 2π/T (the coefficient of t in the position equation):

ω^2 = (2π/T)^2

To find T, we can compare the given position equation with the general form of simple harmonic motion (Y(t) = A sin(ωt + φ)). The coefficient of t in the position equation gives us 2π/T, so:

2π/T = 2π rad

Therefore, T = 1 s.

Now we can calculate ω:

ω = 2π/T = 2π rad/s

Next, calculate the spring constant (k) using Hooke's Law:

k = ma/x

The mass (m) is given as 1.3 kg, and the acceleration (a) is the maximum acceleration, which we'll calculate in the next step.

Step 3: Calculate the maximum acceleration (a)
The maximum acceleration (a) can be found by taking the derivative of the position equation with respect to time (t):

a(t) = d^2Y(t)/dt^2

Plugging in the given position equation Y(t) = (0.09 m) sin(2πt/T + π/3), we differentiate it twice with respect to t to obtain:

a(t) = - ω^2Y(t)

Substituting the values of ω and Y(t):

a(t) = - (2π rad/s)^2 * (0.09 m) sin(2πt/1 + π/3)

To find the maximum acceleration, we need to take the absolute value of this equation and evaluate it for t = 0:

|a(0)| = (2π rad/s)^2 * (0.09 m) sin(π/3)

Step 4: Calculate the potential energy (U)
Plugging in the values of x and k in the potential energy equation:

U = (1/2)kx^2

Note that we have already calculated k in step 2. So substituting the values:

U = (1/2)(k)(0.21 m)^2

Step 5: Calculate the kinetic energy (K)
The kinetic energy (K) at any given time can be calculated as:

K = (1/2)mv^2

where m is the mass of the object and v is its velocity.

In this case, we need to find the maximum speed, which occurs when the displacement from equilibrium is maximum. When the displacement is maximum, the velocity is zero, so the kinetic energy is also zero.

Step 6: Calculate the total mechanical energy (E)
The total mechanical energy (E) is given by the sum of the potential energy (U) and kinetic energy (K):

E = U + K

For t = 0 s, both U and K are at their maximum values. Therefore, the total mechanical energy at t = 0 s is:

E = U + K = U + 0 = U

Now, plug in the values of U to find the mechanical energy of the system relative to the equilibrium position at t = 0 s.

Note: Please provide the values for k and |a(0)| calculated in steps 2 and 3 respectively in order to find the final result for the mechanical energy.