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.

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answered: 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 $x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)$ .

answered: Guest

i think its

answered: Guest

grav potential energy

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