A 0.50-kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.15 m. Determine (a) the velocity when it passes the equilibrium point, (b) the velocity when it is 0.10 m from equilibrium, (c) the total energy of the system, and (d) the equation describing the motion of the mass, assuming that at t = 0, x was a maximum.

Here some facts about simple harmonic motion to help you answer these yourself.

(a) The angular frequency is w = 2 pi f radians per second.
Tha maximum velocity is w A, where a is the amplitude.
The spring constant k is related to w and M by
w = sqrt(k/m)
Compute w. You will need it later.

(b) 0.10 m is 2/3 of the maximum deflection. Potential energy is proportion do the square of deflection, and will be (2/3)^2 = 4/9 of the maximum energy, or
(4/9)*(1/2)*k*(Amplitude)^2. The velocity will be (5/9) of the maxiomum energy. Use that to solve for velocity.
(c) I already mentioned that the toal energy is (1/2)*k*(Amplitude)^2
(d) X = A cos wt

(a) 2,83ms (b) 2,11ms (c) 2,00j (d) x=acoswt .

To solve this problem, we'll use the equations for simple harmonic motion:

1. The angular frequency, ω, is given by ω = 2πf, where f is the frequency of vibration.
2. The maximum velocity, vmax, is given by vmax = Aω, where A is the amplitude of oscillation.
3. The total mechanical energy, E, is given by E = (1/2)kA^2, where k is the spring constant.

Given:
Mass, m = 0.50 kg
Frequency, f = 3.0 Hz
Amplitude, A = 0.15 m
Distance from equilibrium, x = 0.10 m

(a) Velocity when it passes the equilibrium point:
ω = 2πf = 2π(3.0) = 6π rad/s

v = vmax at equilibrium = Aω = (0.15)(6π) ≈ 2.83 m/s

(b) Velocity when it is 0.10 m from equilibrium:
At any point x from equilibrium, the velocity v can be given by v = ω√(A^2 - x^2). Here, x = 0.10 m.

v = ω√(A^2 - x^2) = (6π)√(0.15^2 - 0.10^2) ≈ 3.91 m/s

(c) Total energy of the system:
E = (1/2)kA^2

To find k, we can use the formula: ω^2 = k/m

k = ω^2m = (6π)^2 * 0.50 ≈ 565.49 N/m

E = (1/2)kA^2 = (1/2)(565.49)(0.15^2) ≈ 7.98 J

(d) Equation describing the motion of the mass:
The equation describing simple harmonic motion is given by:

x(t) = A sin(ωt + φ), where φ is the phase constant.

Since at t = 0, x was a maximum, we know that φ = 0.

Therefore, the equation describing the motion is:

x(t) = 0.15 sin(6πt)

Please note that the equation is in terms of displacement, x, and time, t.

To answer the given questions, we need to apply the principles of simple harmonic motion. We'll use the formulae related to the velocity, displacement, and energy of a mass-spring system.

(a) To determine the velocity when the mass passes the equilibrium point, we can use the formula:

v = ω * A

where v represents the velocity, ω is the angular frequency, and A is the amplitude.

First, we need to find the angular frequency ω. The angular frequency is given by the formula:

ω = 2πf

where ω represents the angular frequency and f is the frequency.

Given the frequency f = 3.0 Hz, we can calculate the angular frequency:

ω = 2π * 3.0 = 6π rad/s

Now, we can calculate the velocity:

v = (6π rad/s) * 0.15 m = 0.9π m/s

Therefore, the velocity when the mass passes the equilibrium point is 0.9π m/s.

(b) To find the velocity when the mass is 0.10 m from equilibrium, we can use the formula for velocity at any displacement x from equilibrium:

v = ω * √(A^2 - x^2)

where v represents the velocity, ω is the angular frequency, A is the amplitude, and x is the displacement.

Plugging in the given values, we get:

v = (6π rad/s) * √[(0.15 m)^2 - (0.10 m)^2 ]

v ≈ (6π rad/s) * √(0.0225 m^2 - 0.01 m^2)

v ≈ (6π rad/s) * √(0.0125 m^2)

v ≈ (6π rad/s) * 0.1118 m

v ≈ 0.67π m/s

Therefore, the velocity when the mass is 0.10 m from equilibrium is approximately 0.67π m/s.

(c) To determine the total energy of the system, we need to consider both the kinetic energy (KE) and potential energy (PE).

The total energy (E) of a mass-spring system can be found using the formula:

E = KE + PE

The kinetic energy is given by the formula:

KE = (1/2) * m * v^2

where m represents the mass, and v represents the velocity.

The potential energy is given by the formula:

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

where k represents the spring constant, and x represents the displacement.

Given that m = 0.50 kg, v = 0.67π m/s (from part b), A = 0.15 m, and x = 0.10 m, we can calculate the total energy.

First, we need to find the spring constant (k). The formula to calculate the spring constant for a mass-spring system is:

k = (4π^2) * m * f^2

Plugging in the values, we get:

k = (4π^2) * 0.50 kg * (3.0 Hz)^2

k = (4π^2) * 0.50 kg * 9.0 s^-2

k ≈ (4π^2) * 0.50 kg * 9.0

k ≈ (4π^2) * 4.5 kg

k ≈ 56.5 N/m (rounded to one decimal place)

Now, we can find the total energy:

KE = (1/2) * (0.50 kg) * (0.67π m/s)^2

KE ≈ (1/2) * (0.50 kg) * (0.67π m/s)^2

KE ≈ 0.034π^2 J (rounded to three decimal places)

PE = (1/2) * (56.5 N/m) * (0.10 m)^2

PE ≈ (1/2) * (56.5 N/m) * (0.10 m)^2

PE ≈ 0.142 J

E = KE + PE

E ≈ 0.034π^2 J + 0.142 J

E ≈ 0.034π^2 J + 0.142 J

E ≈ 0.034π^2 J + 0.142 J

E ≈ 0.142 J + 0.034π^2 J

E ≈ 0.307 J (rounded to three decimal places)

Therefore, the total energy of the system is approximately 0.307 J.

(d) To find the equation describing the motion of the mass, assuming that at t = 0, x was a maximum, we can use the equation:

x = A * cos(ωt + φ)

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

Given that A = 0.15 m and ω = 6π rad/s, we need to find the phase constant φ.

At t = 0, x is a maximum, which means cos(ωt + φ) = 1.

Therefore, we need to solve the equation: A * cos(φ) = 1

0.15 * cos(φ) = 1

cos(φ) = 1 / 0.15

φ = acos(1 / 0.15)

Using a calculator, we find φ ≈ 1.533.

Now, we can write the equation describing the motion of the mass:

x = (0.15 m) * cos(6πt + 1.533)

Therefore, the equation describing the motion of the mass is x = (0.15 m) * cos(6πt + 1.533).