Suppose that the force acting on a mass m is F=-kx+fsin(omega)t.

a.) show that x(t)= A sin omegat is a possible motion for this simple driven oscillator if A= f/k-m omega^2

b.) what is the amplitude when the frequency is very small?

I hav no idea where to start this

To solve this problem, we'll start by applying Newton's second law to the given system. Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the force acting on the mass m is given by F = -kx + fsin(ωt), where k is the spring constant, x is the displacement of the mass from its equilibrium position, f is the driving amplitude, ω is the driving frequency, and t is time.

Applying Newton's second law, we have:
ma = -kx + fsin(ωt)

Next, we need to express acceleration, a, in terms of displacement, x, and time, t. The second derivative of displacement with respect to time gives us acceleration. So, we differentiate x(t) = A sin(ωt) twice with respect to time to get the second derivative.

x(t) = A sin(ωt)
Differentiating once with respect to time:
v(t) = Aω cos(ωt)
Differentiating again:
a(t) = -Aω^2 sin(ωt)

Now, we substitute x(t) and a(t) into the equation ma = -kx + fsin(ωt) to find the conditions under which x(t) is a possible motion.

ma = m(-Aω^2 sin(ωt)) = -k(A sin(ωt)) + f sin(ωt)

Expanding and simplifying, we have:
-mAω^2 sin(ωt) = -kA sin(ωt) + f sin(ωt)

Factoring out sin(ωt), we get:
(-mAω^2 - kA + f) sin(ωt) = 0

To satisfy this equation for all values of t, we must have:
-mAω^2 - kA + f = 0

Rearranging this equation, we get:
f = kA + mAω^2

Now, we can solve for the amplitude A when the frequency ω is very small. When ω is small, sin(ωt) approaches zero, and the driving force term, fsin(ωt), becomes negligible. In this case, the equation reduces to:
f ≈ -kA

Now, solving for A:
A = -f/k

So, when the frequency is very small, the amplitude A is given by A = -f/k.