A 0.26 kg mass is attached to a light spring with a force constant of 36.9 N/m a

Question

A 0.26 kg mass is attached to a light spring with a force constant of 36.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cmand released from rest, determine the following.

(a) maximum speed of the oscillating mass
m/s

(b) speed of the oscillating mass when the spring is compressed 1.5 cm
m/s

(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
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

a)
w = sqrt (k/m)
= sqrt (36.9/0.26)
= 11.9 rad/s

A = 0.05 m

vmax = A*w
= 0.05*11.9
= 0.6 m/s
Answer: 0.6 m/s

b)
use conservation of energy:
Maximum potential energy = potential energy + kinetic energy
0.5*K*A^2 = 0.5*k*x^2 + 0.5*m*v^2
k*A^2 = k*x^2 + m*v^2
36.9*(0.05)^2 = 36.9*(0.015)^2 + 0.26*v^2
v= 0.57 m/s
Answer: 0.57 m/s

c)
compression x = 5 cm - 1.5 cm = 3.5 cm = 0.035 m
Maximum potential energy = potential energy + kinetic energy
0.5*K*A^2 = 0.5*k*x^2 + 0.5*m*v^2
k*A^2 = k*x^2 + m*v^2
36.9*(0.05)^2 = 36.9*(0.035)^2 + 0.26*v^2
v= 0.43 m/s
Answer: 0.43 m/s

d)
v = vmax/2
KE = KEmax/4
PE = 3*PEmax/4

Maximum potential energy = potential energy
3*0.5*K*A^2/4 = 0.5*k*x^2
3*A^2/4 = x^2
x = 0.87*A
= 0.87*5 cm
= 4.33 cm
Answer: 4.33 cm

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