Q. 86 - Calculus

Question

A mass hanging at the end of a spring oscillates up and down from its equilibrium position with velocity

$v\left(t\right)=3\cos \left(\frac{3t}{\sqrt{2}}\right)-3\sqrt{2}\sin \left(\frac{3t}{\sqrt{2}}\right)$

centimetres per second, as shown in the figure. The mass is at its equilibrium at $t=0$. Use definite integrals to determine whether the mass will be above or below its equilibrium position at times $t=4 \mathrm{and} t=5$.

Step-by-Step Solution

Answer

Since the value is in negative form so the mass will be below its equilibrium position at times t = 4 and t = 5.

Step 1. Given Information

A mass hanging at the end of a spring oscillates up and down from its equilibrium position with velocity

$v\left(t\right)=3\cos \left(\frac{3t}{\sqrt{2}}\right)-3\sqrt{2}\sin \left(\frac{3t}{\sqrt{2}}\right)$

centimetres per second, as shown in the figure. The mass is at its equilibrium at $t=0$. Use definite integrals to determine whether the mass will be above or below its equilibrium position at times$t=4 \mathrm{and} t=5$.

Step 2. Using definite integrals to determine whether the mass will be above or below its equilibrium position at times t = 4   and   t = 5 .

$$\begin{gathered} {\int }_4^{5}v\left(t\right)dt={\int }_4^5\left[3\cos \left(\frac{3t}{\sqrt{2}}\right)-3\sqrt{2}\sin \left(\frac{3t}{\sqrt{2}}\right)\right]dt \\ {\int }_4^{5}v\left(t\right)dt=3{\int }_4^5\cos \left(\frac{3t}{\sqrt{2}}\right)dt-3\sqrt{2}{\int }_4^5\sin \left(\frac{3t}{\sqrt{2}}\right)dt \end{gathered}$$

Let

$$\begin{gathered} u=\frac{3t}{\sqrt{2}} \\ \frac{du}{dt}=\frac{3}{\sqrt{2}} \\ du==\frac{3}{\sqrt{2}}dt \\ \frac{\sqrt{2}}{3}du=dt \end{gathered}$$

Step 3. Now the integral is

$$\begin{gathered} {\int }_4^{5}v\left(t\right)dt=3·\frac{\sqrt{2}}{3}{\int }_4^5\cos udu-3\sqrt{2}·\frac{\sqrt{2}}{3}{\int }_4^5\sin udu \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}{\int }_4^5\cos udu-2{\int }_4^5\sin udu \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin u\right]}_4^5-2{[-\cos u]}_4^5 \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5+2{\left[\cos \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5 \end{gathered}$$

Step 5. Now simplifying the integral.

$$\begin{gathered} {\int }_4^{5}v\left(t\right)dt=\sqrt{2}{\left[\sin \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5+2{\left[\cos \left(\frac{3t}{\sqrt{2}}\right)\right]}_4^5 \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(\frac{3\times 5}{\sqrt{2}}\right)-\sin \left(\frac{3\times 4}{\sqrt{2}}\right)\right]+2\left[\cos \left(\frac{3\times 5}{\sqrt{2}}\right)-\cos \left(\frac{3\times 4}{\sqrt{2}}\right)\right] \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(\frac{15}{\sqrt{2}}\right)-\sin \left(\frac{12}{\sqrt{2}}\right)\right]+2\left[\cos \left(\frac{15}{\sqrt{2}}\right)-\cos \left(\frac{12}{\sqrt{2}}\right)\right] \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}\left[\sin \left(10.64\right)-\sin \left(8.51\right)\right]+2\left[\cos \left(10.64\right)-\cos \left(8.51\right)\right] \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}\left[-0.94-0.79\right]+2\left[-0.35-(-0.61)\right] \\ {\int }_4^{5}v\left(t\right)dt=\sqrt{2}\left[-0.94-0.79\right]+2\left[-0.35+0.61\right] \\ {\int }_4^{5}v\left(t\right)dt=1.41\times (-1.73)+2\times 0.26 \\ {\int }_4^{5}v\left(t\right)dt=-2.44+0.52 \\ {\int }_4^{5}v\left(t\right)dt=-1.92 \end{gathered}$$