Question

A 5kg collar is attached to a spring of force constant 500n/m it slides without friction on a horizontal rod the collar is displaced from its equilibrium position by 10.0cm and released calculate the period of oscillation the maximum speed and the maximum acceleration of the collar

Answer 1

m = 5kg

k = 500 Nm^ -1

A = 10 cm = 0.01 m

T = 2 π √ m/k = 2 x 3.14 √ 5 / 500 = 0.6284 s

v_m = A = A √ k / m = 0.1 √ 500 / 5 = 1 ms^-1

a_max = k / m x A = 500 / 5 x 0. 1 = 10 ms^-2

Answer 2

hey frnd ❤️❤️

please refer to the attachment