Question

A trolley of mass 3 kg is connected to two springs having spring constant as 600 each and is displaced from equilibrium position by 5 cm

Answer 1

Hey dear,
The question is incomplete. Complete question is given in image.

◆ Answer-
a) T = 0.314 s
b) vmax = 1 m/s
c) E = 1.5 J

◆ Explanation-
# Given-
k1 = k2 = 600 N/m
m = 3 kg
A = 5 cm = 0.05 m

# Solution-
Combined force constant of spring is calculated by-
k = k1 + k2
k = 600 + 600
k = 1200 N/m

Angular velocity is given by -
w = √(k / m)
w = √(1200/3)
w = 20 rad/s

a) Time period of oscillation is calculated by -
T = 2π / w
T = 2×3.142 / 20
T = 0.314 s

b) Maximum velocity is given by -
v = Aw
v = 0.05 × 20
v = 1 m/s

c) Energy dissipated due to damping forces is calculated by -
E = 1/2 mA^2w^2
E = 1/2 × 3 × 0.05^2 × 20^2
E = 1.5 J

Hope this help you...