A .6-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of .13 m. Determine a.) the velocity when it passes the equilibrium point, b.) the velocity when it is .1 m from equilibrium. c.) the total energy of the system, and d.) the equation describing the motion of the mass, assuming that x was a maximum at t=0?

This is WRONG. Did not multiply the 2pi by 3. You just did 2pi

To solve this problem, we need to use the equations of motion for a mass-spring system. Let's go step-by-step to find the solutions to the given questions:

a.) The velocity when the mass passes the equilibrium point:
At the equilibrium point, the velocity of the mass is maximum. The maximum velocity of an object in simple harmonic motion is given by the equation:
v_max = Aω

where A is the amplitude of motion and ω is the angular frequency.

Given that the amplitude (A) is 0.13 m and the frequency (f) is 3.0 Hz, we can use the equation:
ω = 2πf

Substituting the values into the equation, we get:
ω = 2π(3.0 Hz) = 6π rad/s

Now, we can calculate the maximum velocity:
v_max = Aω = (0.13 m)(6π rad/s) ≈ 0.78π m/s

Therefore, the velocity when the mass passes the equilibrium point is approximately 0.78π m/s.

b.) The velocity when the mass is 0.1 m from equilibrium:
Using the equation for velocity in simple harmonic motion, we can find the velocity when the mass is at a displacement x from the equilibrium point:

v = ±ω√(A² - x²)

Given that the displacement from equilibrium (x) is 0.1 m and the angular frequency (ω) is 6π rad/s, we can substitute these values and calculate the velocity:
v = ±(6π rad/s)√(0.13² m² - 0.1² m²) ≈ ±(6π rad/s)√(0.0169 m² - 0.01 m²)

v ≈ ±(6π rad/s)√(0.0069 m²)

Therefore, the velocity when the mass is 0.1 m from equilibrium is approximately ±0.06π m/s.

c.) The total energy of the system:
The total energy of the system in simple harmonic motion is the sum of the kinetic energy (KE) and the potential energy (PE). Mathematically, it can be represented as:

E_total = KE + PE

Given that the mass (m) is 0.6 kg, the angular frequency (ω) is 6π rad/s, and the amplitude (A) is 0.13 m, we can calculate the total energy:

KE = (1/2)mv², where v is the maximum velocity calculated in part a.
KE = (1/2)(0.6 kg)(0.78π m/s)²

PE = (1/2)kx², where k is the spring constant and x is the amplitude. The spring constant (k) of the spring can be calculated using the formula:
k = mω²

k = (0.6 kg)((6π rad/s)²)

PE = (1/2)(0.6 kg)((6π rad/s)²)(0.13 m)²

Finally, we can substitute the values into the equation for total energy:

E_total = KE + PE

Substituting the calculated values, we get:
E_total = (1/2)(0.6 kg)(0.78π m/s)² + (1/2)(0.6 kg)((6π rad/s)²)(0.13 m)²

Therefore, the total energy of the system can be calculated using these values.

d.) The equation describing the motion of the mass:
The equation describing the motion of the mass can be given by:

x = Acos(ωt + φ)

where x is the displacement from the equilibrium position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

Given that the displacement from equilibrium (x) is maximum at t = 0, we can conclude that φ = 0. Therefore, the equation becomes:

x = Acos(ωt)

Substituting the values of A and ω, we get:
x = (0.13 m)cos((6π rad/s)t)

Hence, the equation describing the motion of the mass is (approximately):
x = 0.13cos(6πt)

To answer these questions, we'll need to use the equations of motion for a mass-spring system.

The equation describing the motion of a mass-spring system is given by:
\[x(t) = A \sin(\omega t + \phi)\]
where:
- \(x(t)\) is the displacement of the mass from the equilibrium position at time \(t\)
- \(A\) is the amplitude of the motion
- \(\omega\) is the angular frequency of the system
- \(\phi\) is the phase constant

The angular frequency \(\omega\) can be calculated using the formula:
\[\omega = 2\pi f\]
where:
- \(f\) is the frequency of the motion

Now, let's use these equations to answer the questions one by one.

a.) To find the velocity when the mass passes the equilibrium point, we need to determine the rate of change of displacement with respect to time, which is the derivative of \(x(t)\) with respect to \(t\).

Taking the derivative of \(x(t)\) with respect to \(t\), we get:
\[\frac{{dx}}{{dt}} = A \omega \cos(\omega t + \phi)\]

When the mass passes the equilibrium point, \(x(t) = 0\), so we can find the velocity by substituting \(x(t) = 0\) into the equation above:
\[\frac{{dx}}{{dt}} = A \omega \cos(\omega t + \phi) = 0\]

Since \(\cos(\omega t + \phi) = 0\) when \(\omega t + \phi = \frac{{\pi}}{2}\) or \(3\frac{{\pi}}{2}\), we have:
\[\omega t + \phi = \frac{{\pi}}{2}\]
or
\[\omega t + \phi = 3\frac{{\pi}}{2}\]

Solving for \(t\) gives us two solutions:
\[\omega t = \frac{{\pi}}{2} - \phi\]
or
\[\omega t = 3\frac{{\pi}}{2} - \phi\]

Therefore, the two times when the mass passes the equilibrium point are:
\[t_1 = \frac{{\frac{{\pi}}{2} - \phi}}{{\omega}}\]
and
\[t_2 = \frac{{3\frac{{\pi}}{2} - \phi}}{{\omega}}\]

To find the velocity at these times, substitute these values into \(\frac{{dx}}{{dt}}\):
\[v_1 = A \omega \cos(\omega t_1 + \phi)\]
\[v_2 = A \omega \cos(\omega t_2 + \phi)\]

b.) To find the velocity when the mass is \(0.1\,m\) from equilibrium, we can substitute \(x(t) = 0.1\) into the equation \(x(t) = A \sin(\omega t + \phi)\) and solve for \(t\).

\[0.1 = A \sin(\omega t + \phi)\]

To find the velocity at this time, substitute this value of \(t\) into \(\frac{{dx}}{{dt}}\):
\[v = A \omega \cos(\omega t + \phi)\]

c.) The total energy of the system is the sum of kinetic energy (\(K\)) and potential energy (\(U\)). In this case, \(K\) is the kinetic energy of the mass given by \(K = \frac{1}{2} m v^2\), and \(U\) is the potential energy of the spring given by \(U = \frac{1}{2} k x^2\), where \(m\) is the mass, \(v\) is the velocity, \(k\) is the spring constant, and \(x\) is the displacement from equilibrium.

d.) The equation describing the motion of the mass is given by \(x(t) = A \sin(\omega t + \phi)\), where we need to determine \(A\), \(\omega\), and \(\phi\) based on the given information.

To find \(A\), use the amplitude given in the problem statement.

To find \(\omega\), use the formula for angular frequency \(\omega = 2\pi f\) and the given frequency.

To find \(\phi\), we need additional information, such as the initial phase angle or the position and velocity at a specific time. Please provide any additional information you have so we can determine the value of \(\phi\) and complete the equation describing the motion.

a) Maximum velocity, which happens at the equilibrium point, is

Vmax = A*w, where w is the angular frequency, in radians per second.
In this case,
w = 3.0 * 2 pi = 6.28 radians/s
Vmax = 0.13m*6.28 rad/s = 0.816 m/s

b) 0.1 m from equilibrium is 0.1/0.13 = 76.92% of maximum deflection, and will have 0.7692^2 = 59.17% of maximum spring potential energy, which is 59.17% of total energy.
1-0.5917 = 40.83% of the energy will be kinetic energy at 0.1 m stretch.
Velocity wil be sqrt(0.4083)= 63.90% of maximum, or 0.521 m/s

c) (1/2) M*Vmax^2 = maximum KE
Do the numbers
d) X = (amplitude)*cos (wt)
= 0.13 cos(6.28 t)
(It could also be negatve)