A glider of mass m is attached to a spring that has a force constant k. The glider is set into simple harmonic motion with amplitude A. Determine the positions (by the magnitude) at which the kinetic and the potential energies of the glider are equal. Use a coordinate system whose center is in the equilibrium position of the glider, where the x-axis is directed to the right, and the point of the spring's attachment to the wall has a negative x-coordinate.

Express your answer in terms of the variables m, k, and A.

I need help on this question. Please give me explanations on how to solve this problem. Thanks!

A/sqrt2

To determine the positions at which the kinetic and potential energies of the glider are equal, we first need to understand the equations for kinetic and potential energy in simple harmonic motion.

The kinetic energy (KE) of an object is given by the equation:
KE = (1/2)mv^2

where m is the mass of the glider and v is its velocity.

The potential energy (PE) of an object attached to a spring is given by the equation:
PE = (1/2)kx^2

where k is the force constant of the spring and x is the displacement of the glider from its equilibrium position.

In simple harmonic motion, the total energy (TE) remains constant, which means that KE + PE is constant.

Therefore, we can set up the equation:
(1/2)mv^2 + (1/2)kx^2 = TE

Since we want to find the positions at which the kinetic and potential energies are equal, we can set KE equal to PE:

(1/2)mv^2 = (1/2)kx^2

Simplifying the equation, we can cancel out the common factors:

mv^2 = kx^2

To solve for the positions (x) at which KE = PE, we can eliminate v from the equation using the relationship between velocity and displacement in simple harmonic motion.

The velocity (v) of an object in simple harmonic motion is given by the equation:
v = ω√(A^2 - x^2)

where ω is the angular frequency and A is the amplitude of the motion.

By substituting this expression for v into our equation, we get:

m(ω√(A^2 - x^2))^2 = kx^2

Simplifying further, we have:

mω^2(A^2 - x^2) = kx^2

Expanding the equation:

mω^2A^2 - mω^2x^2 = kx^2

Rearranging the terms:

(mω^2A^2) = (mω^2 + k)x^2

Finally, we solve for x:

x^2 = (mω^2A^2)/(mω^2 + k)

Taking the square root on both sides:

x = √((mω^2A^2)/(mω^2 + k))

Therefore, the positions at which the kinetic and potential energies of the glider are equal are given by x = √((mω^2A^2)/(mω^2 + k)).

Note: ω^2 = k/m is the expression for the angular frequency of the oscillation.