A 0.94 kg mass is attached to a light spring with a force constant of 42.9 N/m a

Question

A 0.94 kg mass is attached to a light spring with a force constant of 42.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched S.0 cm and reieased from rest, determine the following ) maximum speed of the oscillating mass m/s (b) speed of the oscilating mass when the spring is compressed 1.5 crm nv's (C) speed of the oscillating mass as it passes the point 1,5 ? from the equlibrium position nys 12 value of x at which the speed of the osalleting mass is equal to one-haif the maximum value Additional Materials

Explanation / Answer

a)

Maximum Speed

Vmax=A*sqrt(K/m) =0.05*sqrt(42.9/0.94)

Vmax=0.3378 m/s

b)

By Conservation of energy

(1/2)Kx2+(1/2)mV2 =(1/2)KA2

42.9*0.0152+0.94*V2 =42.9*0.052

V=0.3222 m/s

C)

V=0.3222 m/s

d)

Now V=0.3378/2=0.16889 m/s

By Conservation of energy

(1/2)Kx2+(1/2)mV2 =(1/2)KA2

42.9*x2+0.94*0.168892 =42.9*0.052

X=0.0433 m

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