Respuesta :
Answer:
A)
0.395 m
B)
2.4 m/s
Explanation:
A)
[tex]m[/tex] = mass of the cart = 1.4 kg
[tex]k[/tex] = spring constant of the spring = 50 Nm⁻¹
[tex]x[/tex] = initial position of spring from equilibrium position = 0.21 m
[tex]v_{i}[/tex] = initial speed of the cart = 2.0 ms⁻¹
[tex]A[/tex] = amplitude of the oscillation = ?
Using conservation of energy
Final spring energy = initial kinetic energy + initial spring energy
[tex](0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m[/tex]
B)
[tex]m[/tex] = mass of the cart = 1.4 kg
[tex]k[/tex] = spring constant of the spring = 50 Nm⁻¹
[tex]A[/tex] = amplitude of the oscillation = 0.395 m
[tex]v_{o}[/tex] = maximum speed at the equilibrium position
Using conservation of energy
Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring
[tex](0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}[/tex]
Answer:
(a) 0.4 m
(b) 2.392 m/s
Explanation:
mass, m = 1.4 kg
spring constant, k = 50 N/m
x = 0.21 m
velocity maximum, v = 2 m/s
(A) Angular frequency, [tex]\omega =\sqrt{\frac{K}{m}}[/tex]
[tex]\omega =\sqrt{\frac{50}{1.4}}[/tex]
ω = 5.98 rad/s
(a) Let amplitude is A.
[tex]v=\omega \sqrt{A^{2}-x^{2}}[/tex]
[tex]2=5.98 \sqrt{A^{2}-0.21^{2}}[/tex]
0.1119 = A² - 0.0441
A = 0.4 m
(b) At the equilibrium position, the speed is maximum
maximum speed, v = ωA = 5.98 x 0.4 = 2.392 m/s
