Amass of 0.40 kg, hanging from a spring with a spring constant of 80 n/m, is set into an up-and-down simple harmonic motion. what is the speed of the mass when moving through the equilibrium point? the starting displacement from equilibrium is 0.10 m.

v = 1.22 m/s

Explanation:

Given that,

Mass of the object, m = 0.4 kg

Spring constant of the spring, k = 80 N/m

The starting displacement from equilibrium is 0.10 m, or the amplitude, A = 0.1 m

To find,

The speed of the mass when moving through the equilibrium point.

Displacement in the mass, y = 0.05

Solution,

Let is the angular frequency of the mass. It can be calculated as :

When the mass displaces from some position, the speed of the mass is given by :

v = 1.22 m/s

So, the speed of the mass when moving through the equilibrium point is 1.22 m/s.

i think its c if i'm wrong

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015