# A mass suspended on a spring will exhibit sinusoidal motion when it moves. If the mass on a spring is 85 cm off the ground at its highest position and 41 cm off the ground at its lowest position...

A mass suspended on a spring will exhibit sinusoidal motion when it moves. If the mass on a spring is 85 cm off the ground at its highest position and 41 cm off the ground at its lowest position and takes 3.0 s to go from the top to the bottom and back again, determine an equation to model the data.

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### 1 Answer

Answer: `y(t) = 22text(cm)*cos(2.1text(Hz)*t)+63text(cm)` * If you need the answer in terms of `pi` , then read the Approach/Solution section.

**Background and Plan:**

Here is a basic sinusoidal.

The blue line represents the period, `T` . The amount of time for a complete wavelength to occur.

To make things simple, we will choose the cosine curve to model our data. The reason: cos(0) = 1, which represents a maximum value.

`phi=0` : (Phase shift = 0) Typically the graph can be shifted left or right, again, we will assume that we release the mass at the maximum and start timing from there. Thus we take the generic form of the cosine and simplify it:

`y(t) = Acos(omegat+phi)+y_0 to Acos(omegat)+y_0`

Where y_0 is the equilibrium point, and `omega` is the angular frequency. Angular frequency is defined as , `f` is frequency. Since `T=1/f` we put it all together:

`y(t) = Acos((2pi)/Tt)+y_0`

Now we need ti fill in the values for `y_0` , `T` and `A` from the problem statement.

**Approach and Solution:**

Using the equation as a guide, determine the equilibrium point, amplitude, and the period from the given information.

1. The equilibrium point is halfway between the minimum and maximum, so take the average of those two given numbers to get `y_0` .

`y_0=(y_(max)+y_(min))/2 = 63 cm`

2. Amplitude is the height above or below the equilibrium point, so take the difference between the maximum and the equilibrium point to get A.

`A = y_max - y_0 or y_0-y_min = 22 cm`

3. T was given directly in the problem since it returns to its original spot

T = 3.0 s

Put it all together:

`y(t) = 22text(cm)*cos((2pi)/(3.0text(s))t)+63text(cm)`

Which can be approximated by:

`y(t) = 22text(cm)*cos(2.1text(Hz)*t)+63text(cm)`

**Evaluate**

Here's the graph for six seconds:

Does it match what is given in the problem statement?

ANSWER: `y(t) = 22text(cm)*cos(2.1text(Hz)*t)+63text(cm)`

**Sources:**