Question

A mass of 1.25 kg stretches a spring 0.08 mm. The mass is in a medium that exerts a viscous resistance of 120 NN when the mass has a velocity of 6 msms. The viscous resistance is proportional to the speed of the object.

Suppose the object is displaced an additional 0.07 mm and released.

Find an function to express the object's displacement from the spring's equilibrium position, in mm after tt seconds. Let positive displacements indicate a stretched spring, and use 9.8 ms2ms2 as the acceleration due to gravity.

u(t) =

Answer 1

for equilibrium position => spering force = gravitational force K(n) mo x 10.084 103) 1.254 9.8 0.08 Natural length. Equilibrium position. 153125 Neaton. Amplitude (inidial) of motion 0.07 Inom m Ao u 0.07 mm : 7x105m Amplitude Position. (Extreme) m= 1.25 lei viscus resistance borce tu br where is constant - 120 -b (6) b 20 N/mis Now Force – 14 (al-bu ma² dt? - 140 - bda dt -) man dt? + bdn det + Kn this is differensial equation solution of this lequatich So x = Where wa -) - (2) (no ess) sin ( wt+4) *)-( 30.05) m 2 153125 1.25 349.9 rad/sec. d - 1/2 ble at t=0 x is maximum (at extreme) So put values in ern (1)

- 204 x = 0.07 le 2x 1.25 Sin ( 343.5+ + 712) x = 0.07 sin (3 349.9t + 1/2) mm 0.07188t) 1005 lést) sin ( 343.9+ +0,2) xr 7x105 meter dibber in units.