Respuesta :
Answer:
a) x_max = 0.20794 m
b) v_max = 3.8436 m/s
c) P = 0.05883 W
Explanation:
Given:
- The stiffness k = 205 N / m
- The mass m = 0.6 kg
- initial compression of the spring xi = 13 cm
- initial speed of the mass vi = 3 m/s
Find:
(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Solution:
- Conservation of energy principle can be applied that the total energy U of the system remains constant. So the Total energy is:
U = K.E + P.E
U = 0.5*m*v^2 + 0.5*k*x^2
- We will take initial point with given values and maximum compression x_max when v = 0.
0.5*m*vi^2 + 0.5*k*xi^2 = 0.5*k*x_max^2
(m/k)*vi^2 + xi^2 = x_max^2
x_max = sqrt ( (m/k)*vi^2 + xi^2 ) = sqrt ( (.6/205)*3^2 + .13^2
x_max = 0.20794 m
- The angular speed w of the harmonic oscillation is given by:
w = sqrt ( k / m )
w = sqrt ( 205 / 0.6 )
w = 18.48422 rad/s
- The maximum velocity v_max is given by:
v_max = - w*x_max
v_max = - (18.48422)*(0.20794)
v_max = 3.8436 m/s
- The amount of power required to stabilize each oscillation is given by:
P = E_cycle / T
Where, E = Energy per cycle = 0.02 J
T = Time period of oscillation
T = 2π/w
P = E_cycle*w / 2π
P = (0.02*18.48422) / 2π
P = 0.05883 W
For the spring mass system the value of maximum stretch, maximum speed and average power input is 0.20794 m, 3.8436 m/s and 0.581 watt.
What is spring constant?
Spring constant is nothing but the stiffness of the spring. Spring constant is the ratio of force applied on the spring to the displacement of the spring.
The Kinetic energy of spring using spring constant can be given as,
[tex]KE=\dfrac{1}{2}kx^2[/tex]
Here,(k) is the spring constant and (x) is the displacement of the spring.
A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg.
The system released, with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.
- (a) The maximum stretch during the motion-
Total energy of the spring is equal to the sum of KE and PE. Therefore,
[tex]E=KE+PE[/tex]
The total energy of the system is at maximum stretch where, velocity is zero. Therefore,
[tex]\dfrac{1}{2}kx_{max}^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\[/tex]
As, the mass of the spring is 0.6 kg and spring stiffness is 205 N/m. Therefore, put the values,
[tex]\dfrac{1}{2}(205)x_{max}^2=\dfrac{1}{2}(0.6)(3)^2+\dfrac{1}{2}(205)(3)^2\\x_{max}=0.20794\rm m[/tex]
- (b) The maximum speed during the motion-
The maximum speed during the motion can be given as,
[tex]v_{max}=-\dfrac{k}{m}x_{max}[/tex]
Put the values in the above formula as,
[tex]v_{max}=-\sqrt{\dfrac{205}{0.6}}(0.20794)\\v_{max}=-3.8436\rm m/s[/tex]
- (c)The average power input in watts required to maintain a steady oscillation-
The power input for spring mass system can be given as,
[tex]P=\dfrac{E}{2}\sqrt\dfrac{k}{m}[/tex]
For this case, there is energy dissipation of 0.02 J per cycle of the spring-mass system. Thus put the values as,
[tex]P=\dfrac{0.02}{2\pi}\sqrt{\dfrac{205}{0.6}}\\P=0.581\rm W[/tex]
For the spring mass system the value of maximum stretch, maximum speed and average power input is
- (a) The maximum stretch during the motion is 0.20794 meters.
- (b) The maximum speed during the motion is 3.8436 m/s.
- (c) The average power input in watts required to maintain a steady oscillation is 0.581 watt.
Learn more about the spring constant here;
https://brainly.com/question/3510901
