Respuesta :
Answer:
a) The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] 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 [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
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 ([tex]x(t)[/tex]), measured in meters, is:
[tex]x(t) = A\cdot \sin(\omega \cdot t +\phi)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in meters.
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The kinematic equation for the velocity formula of the system ([tex]v(t)[/tex]), measured in meters per second, is derived from (1) by deriving it in time:
[tex]v(t) = \omega\cdot A\cdot \cos (\omega\cdot t+\phi)[/tex] (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:
[tex]v = \pm \omega\cdot A[/tex] (3)
The angular frequency is calculated by this expression:
[tex]\omega = 2\pi\cdot f[/tex] (4)
Where [tex]f[/tex] is the frequency, measured in hertz.
If we know that [tex]f = 3\,hz[/tex] and [tex]A = 0.13\,m[/tex], then the velocity when it passes the equilibrium point, which is the maximum and minimum velocities of the mass:
[tex]\omega = 2\pi\cdot (3\,hz)[/tex]
[tex]\omega \approx 18.850\,\frac{rad}{s}[/tex]
[tex]v = \pm \left(18.850\,\frac{rad}{s} \right)\cdot (0.13\,m)[/tex]
[tex]v = \pm 2.451\,\frac{m}{s}[/tex]
The velocity when it passes the equilibrium point is [tex]\pm 2.451[/tex] meters per second.
b) First, we need to determine the spring constant of the system ([tex]k[/tex]), measured in newtons per meter, in terms of the angular frequency ([tex]\omega[/tex]), measured in radians per second, and mass ([tex]m[/tex]), measured in kilograms. That is:
[tex]k = \omega^{2}\cdot m[/tex] (5)
If we know that [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]m = 0.60\,kg[/tex], then the spring constant is:
[tex]k = \left(18.850\,\frac{rad}{s} \right)^{2}\cdot (0.60\,kg)[/tex]
[tex]k = 213.194\,\frac{N}{m}[/tex]
Lastly, we determine the velocity when the mass is 0.10 meters from equilibrium by the Principle of Energy Conservation:
[tex]U_{k} + K = K_{max}[/tex] (6)
[tex]\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}[/tex] (7)
Where:
[tex]U_{k}[/tex] - Current elastic potential energy, measured in joules.
[tex]K[/tex] - Current translational kinetic energy, measured in joules.
[tex]K_{max}[/tex] - Maximum translational kinetic energy, measured in joules.
[tex]v[/tex] - Current velocity of the system, measured in meters per second.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v_{max}[/tex] - Maximum velocity of the system, measured in meters per second.
If we know that [tex]k = 213.194\,\frac{N}{m}[/tex], [tex]x = 0.10\,m[/tex], [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the velocity of the mass-spring system is:
[tex]\frac{k}{m} \cdot x^{2} + v^{2} = v_{max}^{2}[/tex]
[tex]v^{2} = v_{max}^{2}-\frac{k}{m}\cdot x^{2}[/tex]
[tex]v = \sqrt{v_{max}^{2}-\frac{k\cdot x^{2}}{m} }[/tex] (8)
[tex]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} }[/tex]
[tex]v \approx \pm 1.567\,\frac{m}{s}[/tex]
The velocity when it is 0.10 meters from equilibrium is [tex]\pm 1.567[/tex] meters per second.
c) The total energy of the system ([tex]E[/tex]), measured in joules, can be determined by the following expression derived from the Principle of Energy Conservation:
[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex] (9)
If we know that [tex]m = 0.60\,kg[/tex] and [tex]v_{max} = \pm 2.451\,\frac{m}{s}[/tex], then the total energy of the system is:
[tex]E = \frac{1}{2}\cdot (0.60\,kg)\cdot \left(\pm 2.451\,\frac{m}{s}\right)^{2}[/tex]
[tex]E = 1.802\,J[/tex]
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: ([tex]A = 0.13\,m[/tex], [tex]\omega \approx 18.850\,\frac{rad}{s}[/tex] and [tex]\phi = 0.5\pi\,rad[/tex])
From (1) we obtain the resulting formula:
[tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex] (10)
The equation describing the motion of the mass, assuming that initial position is a maximum is [tex]x(t) = 0.13\cdot \sin (18.850\cdot t +0.5\pi)[/tex].
