A body of mass 5.0 kg is suspended by a spring, which stretches 5 cm when the ma

Question

A body of mass 5.0 kg is suspended by a spring, which stretches 5 cm when the mass is attached. It is then displaced downward an addition 10 cm and released. Its position as a function of time is approximately
A) y = 0.10 cos (14t + .1) B) y = 0.10 sin (14t + .5) C) y = 0.05 cos (14t) D) y = 0.10 sin (14t) E) y = 0.10 cos (14t) A body of mass 5.0 kg is suspended by a spring, which stretches 5 cm when the mass is attached. It is then displaced downward an addition 10 cm and released. Its position as a function of time is approximately
A) y = 0.10 cos (14t + .1) B) y = 0.10 sin (14t + .5) C) y = 0.05 cos (14t) D) y = 0.10 sin (14t) E) y = 0.10 cos (14t)
A) y = 0.10 cos (14t + .1) B) y = 0.10 sin (14t + .5) C) y = 0.05 cos (14t) D) y = 0.10 sin (14t) E) y = 0.10 cos (14t)

Explanation / Answer

At equilibrium ,

k*x = mg

At equilibrium ,the spring stretches by 5 cm

So Amplitude(A) = 0.05 m

k = mg/x = (5*9.8)/0.05 = 980 N/m

Since , w =(k/m) = (980/5) = 14

Equation is given by

y = Acos(wt) = 0.05cos(14t)

Answer) option c) y = 0.05 cos (14t)

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