Physics, 30.12.2020 17:10 des6487

A 0.60-kg mass at the end of a spring vibrates 3.0 times per second with and amplitude of 0.13m. 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 the x was a maximum at t = 0.

Answers: 2

Show answers

Answers

Answer from: petertester7up

a) The velocity when it passes the equilibrium point is $\pm 2.451$ meters per second.

b) The velocity when it is 0.10 meters from equilibrium is $\pm 1.567$ meters per second.

c) The total energy of the system is 1.802 joules.

d) The equation describing the motion of the mass, assuming that initial position is a maximum is $x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)$ .

Explanation:

a) If all non-conservative forces can be neglected and spring has no mass, then the mass-spring system exhibits a simple harmonic motion (SHM). The kinematic formula for the position of the system ( x(t) ), measured in meters, is:

$x(t) = A\cdot \sin(\omega \cdot t +\phi)$ (1)

Where:

- Amplitude, measured in meters.

$\omega$ - Angular frequency, measured in radians per second.

- Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The kinematic equation for the velocity formula of the system ( v(t) ), measured in meters per second, is derived from (1) by deriving it in time:

$v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)$ (2)

The velocity when it passes the equilibrium point occurs when the cosine function is equal to 1 or -1. Then, that velocity is determined by following formula:

$v = \pm \omega\cdot A$ (3)

The angular frequency is calculated by this expression:

$\omega = 2\pi\cdot f$ (4)

Where is the frequency, measured in hertz.

If we know that $f = 3\,hz$ and $A = 0.13\,m$ , then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:

$\omega = 2\pi\cdot (3\,hz)$

$\omega \approx 18.850\,\frac{rad}{s}$

$v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)$

$v = \pm 2.451\,\frac{m}{s}$

The velocity when it passes the equilibrium point is $\pm 2.451$ meters per second.

b) First, we need to determine the spring constant of the system (), measured in newtons per meter, in terms of the angular frequency ( $\omega$ ), measured in radians per second, and mass (), measured in kilograms. That is:

$k = \omega^{2}\cdot m$ (5)

If we know that $\omega \approx 18.850\,\frac{rad}{s}$ and $m = 0.60\,kg$ , then the spring constant is:

$k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)$

$k = 213.194\,\frac{N}{m}$

Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:

$U_{k} + K = K_{max}$ (6)

$\frac{1}{2}\cdot k\cdot x^{2} + \frac{1}{2}\cdot m\cdot v^{2} = \frac{1}{2}\cdot m\cdot v_{max}^{2}$ (7)

Where:

$U_{k}$ - Current elastic potential energy, measured in joules.

- Current translational kinetic energy, measured in joules.

$K_{max}$ - Maximum translational kinetic energy, measured in joules.

- Current velocity of the system, measured in meters per second.

- Mass, measured in kilograms.

$v_{max}$ - Maximum velocity of the system, measured in meters per second.

If we know that $k = 213.194\,\frac{N}{m}$ , $x = 0.10\,m$ , $m = 0.60\,kg$ and $v_{max} = \pm 2.451\,\frac{m}{s}$ , then the velocity of the mass-spring system is:

$\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}$

$v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}$

$v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }$ (8)

$v = \sqrt{\left(\pm 2.451\,\frac{m}{s} \right)^{2}-\frac{\left(213.194\,\frac{N}{m} \right)\cdot (0.10\,m)^{2}}{0.60\,kg} }$

$v \approx \pm 1.567\,\frac{m}{s}$

The velocity when it is 0.10 meters from equilibrium is $\pm 1.567$ meters per second.

c) The total energy of the system (), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:

$E = \frac{1}{2}\cdot m\cdot v_{max}^{2}$ (9)

If we know that $m = 0.60\,kg$ and $v_{max} = \pm 2.451\,\frac{m}{s}$ , then the total energy of the system is:

$E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}$

$E = 1.802\,J$

The total energy of the system is 1.802 joules.

d) Given that initial position of the mass-spring system is a maximum, then we conclude that the equation of motion has the following parameters: ( $A = 0.13\,m$ , $\omega \approx 18.850\,\frac{rad}{s}$ and $\phi = 0.5\pi\,rad$ )

From (1) we obtain the resulting formula:

$x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)$ (10)

The equation describing the motion of the mass, assuming that initial position is a maximum is $x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)$ .

Answer from: Quest

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

Answer from: Quest

prolly a beacuse the momentuem the car has increases while going down hill

Another question on Physics

Physics, 22.06.2019 15:10

Suppose that f : rn → rm and that a ∈ k, where k is a connected subset of rn . suppose further that for each x ∈ k there exists a δx > 0 such that f(x) = f(y) for all y ∈ bδx (x). prove that f is constant on k; that is, f(x) = f(a) for all x ∈ k

Answers: 1

Answer

Physics, 22.06.2019 16:00

What part of the ear is names after tools, such as the hammer and the anvil?

Answers: 1

Answer

Physics, 22.06.2019 21:00

The first law of thermodynamics states that heat added to a system is neither created nor destroyed but is as it changes into other forms of energy.

Answers: 1

Answer

Physics, 22.06.2019 23:40

Acar jack with a mechanical advantage of 6 needs to produce an output force of 600 n to raise a car. what input force is required? 6 n 100 n 600 n 3,600 n

Answers: 2

Answer

You know the right answer?

A 0.60-kg mass at the end of a spring vibrates 3.0 times per second with and amplitude of 0.13m. Det...

Questions