This question makes it difficult to visualize, but thankfully, it does make it easy to use equations to determine what it is going for.

We'll start by determining the governing equation for the system before the larger mass separates by using Hooke's law and Newton's second law:

`F = ma = m ddotx = -kx`

Rearranging, we get a second order homogeneous differential equation:

`ddotx +k/mx = 0`

Substituting known values:

`ddotx+ 100/27.2x = 0`

The commonly-known solution to this problem is the following (assuming no phase delay):

`x(t) = Acos(sqrt(k/m)t)`

We are given the amplitude, so we can get the equation governing our system before mass separation (note: 7 cm = 0.07 m):

`x(t) = 0.07cos(1.92t)`

To determine the point at which acceleration is greatest, we can think intuitively about where acceleration will be greatest, but let's do it mathematically to have the greatest confirmation.

`v = (dx)/(dt) = -0.1344sin(1.92t)`

`a = (dv)/(dt)=-0.2580cos(1.92t)`

`j = (da)/(dt) = 0.4955sin(1.92t)`

Jerk, as the derivative of acceleration, will give us minima and maxima of acceleration. Considering the negative cosine without a phase shift, it is clear the first maximum will be where 1.92*t** = `pi` *. Therefore, *t* = 1.64 sec. This time also happens to give us the maximum value for x(t)! So, we can find the distance and velocity that the system is moving at when the 17 kg mass is removed.

Let's start with velocity. If *t* = 1.64, 1.92*t* = `pi`, and sin(1.92*t*) = 0. Therefore, our velocity at this distance is zero. For the same reasons, except as applied to the cosine in *x*(*t*), our position will simply be *x* = -0.07 m = -7 cm.

So, we have established the same amplitude with a new mass with the following new governing equation:

`x_1(t) = 0.07cos(sqrt(100/10.2)t) = 0.07cos(3.1t)`

Therefore, when we remove the 17 kg mass, we have effectively not changed anything about the amplitude of the system, only the frequency. **The amplitude remains 7 cm.**

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