Simple Harmonic motion problems

[ness9660]

http://sweb.uky.edu/~lsadam0/problem2.JPG


Im kinda lost on this problem. Shouldnt the Y motion be described as A*sin(wt)?

In there equation we are considering when x=0, so the kx term is removed, but where is this third term coming from, and then should Y motion just be described as Asin(-wt-o)?

 

[dextercioby]

Nope,unless,at t=0 the amplitude is zero.Nope,the III-rd term is called "initial phase" and should not be missing when writing the general solution.

Daniel.

 

[thepatientmental]

Thank you for sharing this problem. Simple harmonic motion problems can be tricky, but with some practice and understanding of the concepts, they can be easily solved.

First, let's break down the given equation for x(t):

x(t) = A*cos(wt + ϕ) + C

We can see that the amplitude (A) and the angular frequency (w) are given. The phase angle (ϕ) and the constant (C) are yet to be determined.

Now, looking at the problem, we are given the initial conditions: x(0) = 0 and v(0) = 0.

Plugging in t=0 in the given equation, we get:

x(0) = A*cos(0 + ϕ) + C

Since x(0) = 0, we can conclude that C = 0.

Next, we need to determine the phase angle (ϕ).

Using the given initial velocity (v(0) = 0), we can find ϕ by differentiating x(t) with respect to time:

v(t) = dx(t)/dt = -A*w*sin(wt + ϕ)

Plugging in t=0, we get:

v(0) = -A*w*sin(0 + ϕ) = 0

This means that sin(ϕ) = 0, which gives us ϕ = 0 or ϕ = π.

Now, we can rewrite the equation for x(t) as:

x(t) = A*cos(wt)

And for the y motion, we can use the same equation with a different phase angle:

y(t) = A*cos(wt + π) = -A*cos(wt) = A*sin(wt)

Therefore, the y motion can be described as A*sin(wt).

I hope this explanation helps you understand the problem better. Keep practicing and you will become more comfortable with solving simple harmonic motion problems!