Difference between revisions of "Tutorial Week 10"
From UBCMATH WIKI
Line 29: | Line 29: | ||
<li>$\displaystyle y_p=\frac{2}{\pi(\omega^2-1)}\cos (t)$ (1 mark)</li> | <li>$\displaystyle y_p=\frac{2}{\pi(\omega^2-1)}\cos (t)$ (1 mark)</li> | ||
<li>$y_p=\sum_{n=1}^{10}a_n \cos (nt)$ where $\displaystyle a_n=\frac{2}{n\pi (\omega^2-n^2)}$ (2 marks: 1 for the correct coefficient and 1 for writing the sum correctly)</li> | <li>$y_p=\sum_{n=1}^{10}a_n \cos (nt)$ where $\displaystyle a_n=\frac{2}{n\pi (\omega^2-n^2)}$ (2 marks: 1 for the correct coefficient and 1 for writing the sum correctly)</li> | ||
− | <li> | + | <li> The magnitude of the coefficients decreases as the distance between $n$ and $\omega$ increases. (1 mark)</li> |
</ol> | </ol> | ||
</li> | </li> |
Revision as of 17:11, 23 March 2017
Worksheet Questions
- A undamped mass-spring system with mass 1 kg and spring constant 16 kg/s$^2$, is initially at rest. At $t=3$, a linearly increasing force is applied until the force reaches $F_0 = 10$ N at $t=8$. After that moment, the force remains constant at that level ($F_0$).
- Write down the forcing function for this scenario in terms of Heaviside functions.
- Write down the ODE for this mass-spring system subject to the given forcing function.
- What is the transfer function ($H(s)$) and the impulse response ($h(t)$) for this ODE?
- Use the impulse response and convolution to solve the ODE from part (b).
- Consider the following differential equation:$$ y''+\omega^2 y=\frac{2}{\pi}\cos(t)$$ where $\omega=2.01$.
- Find a particular solution $y_p(t)$ using the method of undetermined coefficients.
- Now consider the equation $$ y''+\omega^2 y=\sum_{n=1}^{10}\frac{2}{n\pi}\cos(nt).$$ Use your solution in (a), and again the method of undetermined coefficients, to determine a particular solution $y_p(t)$.
- Describe how the magnitude of the coefficients in your particular solution varies with n.
Solutions
-
- $f(t)= 2 u_3(t) (t-3) - 2u_8(t)(t-8)$ (2 marks -1 if there is a small error)
- $x''+16x=f(t)$ (1 mark)
- $H(s) = \frac{1}{s^2+16}$ (2 marks) $h(t)=\frac{1}{4} \sin(4t)$ (2 marks -1 if they did not include the coefficient)
- For $t<3$:
- $x(t)=0\text{ (1 mark)}$.
- $x(t) = \int_0^t f(w)h(t-w) \ dw =\frac{1}{2} \int_3^t (w-3) \sin(4(t-w)) \ dw.\text{ (1 mark)}$
- $x(t) =\frac{1}{2} \int_3^8 (w-3) \sin(4(t-w)) \ dw + \frac{1}{2} \int_8^t 5\sin(4(t-w)) \ dw.\text{ (2 marks)}$
-
- $\displaystyle y_p=\frac{2}{\pi(\omega^2-1)}\cos (t)$ (1 mark)
- $y_p=\sum_{n=1}^{10}a_n \cos (nt)$ where $\displaystyle a_n=\frac{2}{n\pi (\omega^2-n^2)}$ (2 marks: 1 for the correct coefficient and 1 for writing the sum correctly)
- The magnitude of the coefficients decreases as the distance between $n$ and $\omega$ increases. (1 mark)