Question

A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the cart is 0.21 m from its equilibrium position, and the initial velocity is 2.0 m/s directed away from the equilibrium position.Part A. What is the amplitude of the oscillation?Part B. What is the speed of the cart at its equilibrium position?

Answer 1

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$x$ = initial position of spring from equilibrium position = 0.21 m

$v_{i}$ = initial speed of the cart = 2.0 ms⁻¹

$A$ = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

$(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m$

B)

$m$ = mass of the cart = 1.4 kg

$k$ = spring constant of the spring = 50 Nm⁻¹

$A$ = amplitude of the oscillation = 0.395 m

$v_{o}$ = 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

$(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}$

Answer 2

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, $\omega =\sqrt{\frac{K}{m}}$

$\omega =\sqrt{\frac{50}{1.4}}$

ω = 5.98 rad/s

(a) Let amplitude is A.

$v=\omega \sqrt{A^{2}-x^{2}}$

$2=5.98 \sqrt{A^{2}-0.21^{2}}$

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