2kg ball in simple harmonic motion

[clh7871]

a 2-kg ball is supended from a spring. when disturbed in a vertical direction, the ball moves up and down in simple harmonic motion with a period of 0.25s.

a. what is the upward force exerted on the ball by the spring when the ball is at the midpoint of its up-and-down path?

b. how much did the spring stretch when the ball was first attached to its end (before the oscillatory motion that started)?

 

[HallsofIvy]

I presume you have, or know how to find (depending upon what level this course is), that x(t)= C cos(&radic(k)t) where x(t) is the height at time t, C is the amplitude and k is the spring constant. The period of such motion is given by √(k)T= 2π which, here, is 0.25 so √(k)= 2π/0.25= 8π and k= 64π².

a. what is the upward force exerted on the ball by the spring when the ball is at the midpoint of its up-and-down path?
That's easy- there is NO total force on the ball at the midpoint so the upward force exerted on the ball by the spring is exactly the downward force exerted by gravity on the ball- its weight.

b. how much did the spring stretch when the ball was first attached to its end (before the oscillatory motion that started)?

You know the weight of the ball and you know the spring constant, k, so use Hooke's law to determine the amount of stretch

 

[wais]

a. At the midpoint of its path, the ball is at equilibrium and therefore the net force acting on it is zero. This means that the upward force exerted by the spring must be equal in magnitude to the downward force of gravity on the ball. Using Newton's second law, we can calculate this force as F = ma, where m is the mass of the ball and a is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we get F = (2 kg)(9.8 m/s^2) = 19.6 N.

b. When the ball was first attached to the spring, it caused the spring to stretch due to the weight of the ball pulling down on it. The amount of stretch can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. In this case, we can use the formula F = kx, where k is the spring constant and x is the distance the spring is stretched. Rearranging the formula, we get x = F/k. Plugging in the values, we get x = (19.6 N)/(k). The amount of stretch will depend on the specific spring constant of the spring used, but for a typical spring with a constant of 10 N/m, the stretch would be x = (19.6 N)/(10 N/m) = 1.96 m.