Question

3. An ideal shock absorption system would use a critically damped oscillator to absorb shock loads. The location of the absorbing piston (𝑥) is described by 𝑥 = 𝜏𝑒−𝛾𝑡 where:

- 𝜏 is the linear damping coefficient

- 𝛾 is the exponential damping constant

- 𝑡 is the time (𝑠)

- 𝑥 is the displacement of piston (𝑚)

The tasks are to:

a) Draw a graph of displacement against time for 𝜏 = 12 and 𝛾 = 2, between 𝑡 = 0𝑠 and 𝑡 = 10𝑠.

b) Calculate the gradient at 𝑡 = 2𝑠 and 𝑡 = 4𝑠.

 QD099_September_2017

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   c) Differentiate the function of 𝑥 and calculate the value of 𝑑𝑥 at 𝑡 = 2𝑠 and 𝑡 = 4𝑠. 𝑑𝑡

d) Compare your answers for part b and part c. (M1)

e) Calculate the derivative for the velocity function(𝑑2𝑥).

Answer 1

Solution

Given that

x = \tau e - \gamma t\

Now for \tau = 12,\,\,\,\gamma = 2\ , we have

x = 12e - 2t\

The plot between $t = 0$ and $t = 10$ is 

Solution (b)

Since the plot is a straight line, therefore, the gradient is constant which is $m=-2$

Henec gradient when $t=2$ s is $m=-2$

And gradient when $t=4$ s is $m=-2$

Solution (c)

x = 12e - 2t\

\frac{{dx}}{{dt}} = 0 - 2(1) = - 2\

Hence $t=2$ s , we have gradient \frac{{dx}}{{dt}} = - 2\

And

Hence $t=4$ s , we have gradient \frac{{dx}}{{dt}} = - 2\

Solution (d)

From (b) and (c), we see that the gradients when $t=2$ s and when $t=4$ s is $m=-2$, fix that is the same.

Solution (e)

x = 12e - 2t\

velocity is v=\frac{{dx}}{{dt}} = 0 - 2(1) = - 2\