Question

A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. The mass passes through the equilibrium point (y=0) with positive velocity at t=0. Positive direction of y-axis is downward.Part AWhat equation describes this motion as a function of time?Part BAt what time will the spring stretch to its maximum length at first time?Part CAt what time will the spring shrink to its minimum length at first time?

Answer 1

Answer:

Explanation:

Amplitude = Y =0 .28 m. Stiffness constant k = 305 N/m ; mass m = .235 kg

angular velocity ω = √ ( k/m) = √ ( 305/ 0.235) = 36 rad / s

A) equation of motion y = Y sin wt =0 .28 sin 36 t

B) for stretching maximum length y = Y

So Y = Y sin 36 t ; 1 = sin 36t ; π/2 = 36t ;

t = 3.14 /(2 x 36 ) = 0.044 s

C ) Time period T = 2π/ω = 0.176 s

In T x 3/4  time it will reach minimum length  position for the first time, so

in time 0.176 x 3/4 =  = 0.132  s it will reach there.