Please help solve the following problem. Thank you! A 0.24 kg mass is attached t

Question

Please help solve the following problem. Thank you!

A 0.24 kg mass is attached to a light spring with a force constant of 27.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass 0.54 m/s (b) speed of the oscillating mass when the spring is compressed 1.5 cm 0.514 m/s (c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position 0.385 When the mass-spring system is oscillating, how does the speed of the oscillating mass when the spring is compressed a certain amount compare to when it is stretched the same amount? m/s (d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value m

Explanation / Answer

Given , m = 0.24 kg , k = 27.9 N/m , A = 0.05 m

a) from energy conservation ,

kA^2/2 = mv^2/2

v = (kA^2/m) = (27.9*0.05^2/0.24)

v = 0.54 m/s

b)at x = 1.5 cm = 0.015 m,

k(A^2 - x^2)/2 = m*v^2/2

So , v = (k(A^2 - x^2)/m)

v = 27.9*(0.05^2 - 0.015^2)/0.24)

v = 0.514 m/s

c) v = 0.514 m/s (speed remains same on both side of the equilibrium position at x = 0.015 m , since the energy is conserved)

d) when v = 0.54/2 = 0.27 m/s

Using conservation of energy ,

k(A^2 - x^2)/2 = m*v^2/2

27.9*(0.05^2 - x^2)/2 = 0.24*0.27^2/2

x^2 = 1.873*10^-3

x = 0.043 m

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