An object mass 1,448 g is attached to a spring with spring constant k = 815 N/m. The object is then displaced by a distance d from the equilibrium position and is released. Shortly after, it passes with speed v = 6.4 m/s through the equilibrium position. By what distance d was the object initially displaced?

i am extremely confused on this question..any help is greatly appreciated!!!

The initial PE was 1/2 kx^2

the equilibrium KE is equal to that

so, 1/2 m 6.4^2=1/2 815*d
put in 1.448kg for m, and solve for d.

use v=sqrt 2* Etotal/m

solve for Etotal

then use Bernoulli's equation and solve for k

To find the initial displacement (d) of the object, we can use the principle of conservation of mechanical energy.

The total mechanical energy of the system consists of two parts: the potential energy stored in the spring and the kinetic energy of the moving object.

The potential energy stored in the spring is given by the formula:
Potential Energy (PE) = (1/2) * k * d^2
where k is the spring constant and d is the displacement from the equilibrium position.

The kinetic energy of the moving object is given by the formula:
Kinetic Energy (KE) = (1/2) * m * v^2
where m is the mass of the object and v is the speed when it passes through the equilibrium position.

Since the total mechanical energy remains constant in the absence of external forces like friction, we can equate the initial potential energy to the final kinetic energy:

(1/2) * k * d^2 = (1/2) * m * v^2

Now we can solve for d:

Rearranging the equation, we get:
d^2 = (m * v^2) / k

Substituting the given values:
d^2 = (1,448 g * (6.4 m/s)^2) / 815 N/m

Converting the mass to kilograms:
d^2 = (1.448 kg * (6.4 m/s)^2) / 815 N/m

Calculating the value of d:
d^2 = 0.0109 m^2
d ≈ 0.1046 m (rounded to four decimal places)

Therefore, the object was initially displaced by approximately 0.1046 meters.