A spring with stiffness of k = 9.5 N/m, lies horizontally on a frictionless tabl

Question

A spring with stiffness of k = 9.5 N/m, lies horizontally on a frictionless table. One end of the spring is attached to a mass of m = 85 g and the other end is attached to the wall. Relaxed length of the spring is L0 = 21 cm. We pull the block to the right, along the positive x-axis, so that length of the spring is now L = 24 cm. At this instant (t = 0) we gently release the block. 1-What is the updated momentum in kg.m/s at t = 0.1 s?

2-What is the updated velocity in m/s at t = 0.1 s?

3-What is the updated position in m at t = 0.1 s?

Explanation / Answer

k = spring constant = 9.5 N/m

m = mass attached to spring = 85 g = 0.085 kg

w = angular frequency = sqrt(k/m) = sqrt(9.5/0.085) = 10.6 rad/s

A = amplitude of shm = L - Lo = 24 - 21 = 3 cm = 0.03 m

equation for the position is given as

x = A Coswt

taking derivative both side

dx/dt = - (Aw) Sinwt

v = - (Aw) Sinwt

velocity at t = 0.1 sec is given as

v = - (0.03 x 10.6) Sin(10.6 x 0.1)

v = - 0.28 m/s

momentum at t = 0.1 sec is given as

P = m v

P = (0.085) (- 0.28) = - 0.024 kgm/s

position at t = 0.1 sec is given as

x = A Coswt

x = (0.03) Cos(10.6 x 0.1)

x = 0.015 m

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