# A mass is oscillating on a spring with a period of 4.60 s. At t=0 s, the mass has zero speed and is at x=8.30 cm.  a) What is the value of t the first time after t=0 s that the mass is at x=4.15 cm? b) What is its acceleration at t= 2.50 s ?

First, we'll start by deriving the equation that describes the motion of the spring. Considering we only have a spring and mass without a damper, the equation is simple:

`x(t) = Acos(omegat- phi)+c`

We will use cosine because we start at the maximum displacement without having to worry about phase...

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First, we'll start by deriving the equation that describes the motion of the spring. Considering we only have a spring and mass without a damper, the equation is simple:

`x(t) = Acos(omegat- phi)+c`

We will use cosine because we start at the maximum displacement without having to worry about phase shift or vertical shift. A can be derived from the maximum displacement, which is given, 0.0830 m. `omega` can be derived from the period with the following relation:

`omega = (2pi)/T`

`omega = (2pi)/4.60`

`omega = 1.37`

So, we have the equation for x(t):

`x(t) = 0.0830cos(1.37t)`

Now, we can easily solve (a) and (b)

a) Find t at which x = 4.15 cm = 0.0415 m

We set up the equation:

`0.0415 = 0.0830cos(1.37t)`

`1/2 = cos(1.37t)`

`pi/6 = 1.37t`

`t = 2.62`

At 2.62 seconds, x will first be 4.15 cm.

b) What is the acceleration at t = 2.50 s?

To do this, we will simply need to take the second derivative of x(t):

`x(t)= 0.0830cos(1.37t)`

`v = (dx)/(dt) = -0.114sin(1.37t)`

`a = (d^2x)/(dt^2) = -0.156cos(1.37t)`

Now, we simply plug in 2.50 for t:

`a = -0.156cos(1.37 * 2.50) = 0.150`

So, our acceleration at time t = 2.50 sec will be 0.150 `m/s^2`.

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