Hello, could anyone help with this excersise of Harmonic Motion?

I have to show that a block of mass "m" that is attached at the top of a frictionless inclined plane using a string of "k" constant , performs simple harmonic motion...
Well doing the the addition of forces that moves the block, I have found that:
mgsen(thetha)-Kx=ma
Then:
gsen(thetha)-(k/m)x=a
And that's all what I have done, I know that aceleration is the second derivative of x, but what happens to "gsen (thetha)" ...
Or is there another way of solve the problem?

g sin theta = constant = your gravitational acceleration for this problem, all it ourg

m a = -kx + m ourg = m d^2x/dt^2

x = Xi + xh sin 2 pi t/T

Xi is our equilibrium x when a = 0
Xi = m ourg/k

now if harmonic then
xh is our harmonic motion x
do it separate as added to Xi

x = xh sin 2 pi t/T
dx/dt = xh(2pi/T) cos 2 pi t/T

d^2x/dt^2 =
-xh(2pi/t)^2 sin 2 pi t/T

so as usual
d^2xh/dt^2 = -(2 pi/T)^2 xh

now do F = m a
-k xh = ma = m (-2 pi /T)^2 xh
sqrt(k/m) = 2 pi/T

T = 2 pi sqrt (m/k)
or
=1/T = (1/2pi) sqrt(k/m)
as usual :)

To show that the block performs simple harmonic motion, we need to show that the equation of motion follows the form of simple harmonic motion, which is a second-order linear differential equation of the form:

d²x/dt² + (ω²)x = 0,

where x is the displacement of the block from its equilibrium position, t is time, and ω is the angular frequency.

In your case, you have the equation:

gsen(theta) - (k/m)x = a,

where g is the acceleration due to gravity, theta is the angle of the inclined plane, k is the spring constant, m is the mass of the block, x is the displacement of the block, and a is the acceleration of the block.

To proceed, let's rewrite the equation in terms of displacement x and the second derivative of x, which is acceleration. Rearranging the equation, we get:

a = gsen(theta) - (k/m)x.

Since a is the second derivative of x, we can rewrite it as:

d²x/dt² = gsen(theta) - (k/m)x.

Now, recall that sin(theta) can also be written as the derivative of cos(theta) with respect to theta. Therefore, we can write sin(theta) as:

sen(theta) = d(cos(theta))/d(theta).

Substituting this back into our equation:

d²x/dt² = g(d(cos(theta))/d(theta)) - (k/m)x.

Now, let's differentiate cos(theta) with respect to theta:

d(cos(theta))/d(theta) = -sin(theta).

Substituting this back into our equation, we get:

d²x/dt² = -g(sin(theta)) - (k/m)x.

Since -g(sin(theta)) is a constant, let's denote it as A:

d²x/dt² = -Ax - (k/m)x.

Now, we can see that our equation follows the form of a simple harmonic motion equation:

d²x/dt² + (k/m)x = 0.

Therefore, you have successfully shown that the motion of the block is simple harmonic motion when subject to the forces you described.

To solve the problem further and find the angular frequency, ω, you can compare the equation above with the general form of simple harmonic motion:

d²x/dt² + (ω²)x = 0.

By comparing the coefficients, you can determine that (k/m) = ω². Taking the square root of both sides, you can find the angular frequency, ω = sqrt(k/m).