The questions refer to a mass m on a horizontal friction-less surface. The mass

Question

The questions refer to a mass m on a horizontal friction-less surface. The mass is attached to one end of a horizontal spring with a spring constant of k. The other end of the spring is fixed in place. Consider the spring to be "mass-less".

1. When the mass is displaced from its equilibrium position by a VECTOR displacement of x, the force of the spring on the mass is

(a.) -0.5kx2

(b.) +kx

(c.) -kx

(d.) none of the given answers is correct

2. When the mass is displaced by x from its equilibrium position, its acceleration is

(a.) -kx/m

(b.) -mx/k

(c.) kx/m

(d.) none of the given answers is correct

3. Suppose you pull the mass to one side so that the spring is stretched, and then let the mass go. When the mass returns to its equilibrium position, it will have (select all that apply)

(a.) 0 velocity

(b.) 0 acceleration

(c.) 0 velocity and 0 accleration

(d.) a net force of zero acting on it

4. After the mass "overshoots" the equilibrium position, it will eventually come to a stop on the other side. At this point, (select all that apply)

(a.) the velocity is 0

(b.) the acceleration is 0

(c.) the displacement from equilibrium is at a maximum

(d.) none of the given answers is correct

5. Consider a mass on a spring oscillating back and forth. When the magnitude of the acceleration is at its greatest value, the (select all that apply)

(a.) displacement from equilibrium is at its greatest

(b.) displacement from equilibrium is 0

(c.) the velocity is 0

(d.) the velocity is at its greatest value

6. For an oscillator (something that moves back and forth), the period of oscillation T is equal to the amount of time for the oscillator to move

(a.) from its equilbrium position to its maximum position and back to equilibrium

(b.) from maximum displacement on one side to the maximum displacement on the other side

(c.) from its equilibrium position to its maximum displacement

(d.) none of the given answers is correct

Explanation / Answer

1. F = - kx

Ans(c)

2. a = F/m = - kx / m

Ans(A)

3. at equilirbium position,

x = 0 so F = -kx = 0

a = F/m = 0

and velocity will be maximum.

Ans(b, d)

4. as it stops v = 0 and x = maximum

Ans: (a, c)

5. (a, c)

6. (d)

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