Question

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.

Answer 1

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)`.

Step-by-step 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:

`A` - Amplitude, measured in meters.

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

`t` - 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
`f` 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\,(rad)/(s)`

`v = \pm \left(18.850\,(rad)/(s) \right)\cdot (0.13\,m)`

`v = \pm 2.451\,(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 (
`k`), measured in newtons per meter, in terms of the angular frequency (
`\omega`), measured in radians per second, and mass (
`m`), measured in kilograms. That is:

`k = \omega^(2)\cdot m` (5)

If we know that
`\omega \approx 18.850\,(rad)/(s)` and
`m = 0.60\,kg`, then the spring constant is:

`k = \left(18.850\,(rad)/(s) \right)^(2)\cdot (0.60\,kg)`

`k = 213.194\,(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)

`(1)/(2)\cdot k\cdot x^(2) + (1)/(2)\cdot m\cdot v^(2) = (1)/(2)\cdot m\cdot v_(max)^(2)` (7)

Where:

`U_(k)` - Current elastic potential energy, measured in joules.

`K` - Current translational kinetic energy, measured in joules.

`K_(max)` - Maximum translational kinetic energy, measured in joules.

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

`m` - Mass, measured in kilograms.

`v_(max)` - Maximum velocity of the system, measured in meters per second.

If we know that
`k = 213.194\,(N)/(m)`,
`x = 0.10\,m`,
`m = 0.60\,kg` and
`v_(max) = \pm 2.451\,(m)/(s)`, then the velocity of the mass-spring system is:

`(k)/(m) \cdot x^(2) + v^(2) = v_(max)^(2)`

`v^(2) = v_(max)^(2)-(k)/(m)\cdot x^(2)`

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

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

`v \approx \pm 1.567\,(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 (
`E`), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:

`E = (1)/(2)\cdot m\cdot v_(max)^(2)` (9)

If we know that
`m = 0.60\,kg` and
`v_(max) = \pm 2.451\,(m)/(s)`, then the total energy of the system is:

`E = (1)/(2)\cdot (0.60\,kg)\cdot \left(\pm 2.451\,(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\,(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)`.