Chapter 10, Problem 1E

Evaluate the following: (a) 5 sin (5t − 9°) at t = 0, 0.01, and 0.1 s; (b) 4 cos 2t at t = 0, 1, and 1.5 s; (c) 3.2 cos (6t + 15°) at t = 0, 0.01, and 0.1 s.

To determine

Find the value of following expression.

Answer to Problem 1E

The value of expression for $t=0\text{s}$ form is $\underset{\_}{-0.78}$, for $t=0.01\text{s}$ form is $\underset{\_}{-0.53}$, and for $t=0.1\text{s}$ form is $\underset{\_}{1.7}$.

Explanation of Solution

Case (i):

Given data:

$5\sin \left(5t-9°\right)$ (1)

Calculation:

Convert the value in degree to radians.

$\begin{array}{c}5\sin \left(5t-9°\right)=5\sin \left(5t-9\left(\frac{\pi }{180}\right)\right)\\ =5\sin \left(5t-9\left(0.017453\right)\right)\end{array}$

$5\sin \left(5t-9°\right)=5\sin \left(5t-0.15707\right)$ (2)

Substitute $0\text{s}$ for $t$ in Equation (2).

$\begin{array}{c}5\sin \left(5t-0.15707\right)=5\sin \left(5\left(0\text{s}\right)-0.15707\right)\\ =5\sin \left(0-0.15707\right)\\ =5\sin \left(-0.15707\right)\\ =5\left(-0.15642\right)\\ =-0.78\end{array}$

Case (ii):

Substitute $0.01\text{s}$ for $t$ in Equation (2).

$\begin{array}{c}5\sin \left(5t-0.15707\right)=5\sin \left(5\left(0.01\text{s}\right)-0.15707\right)\\ =5\sin \left(0.05-0.15707\right)\\ =5\sin \left(-0.10707\right)\\ =5\left(-0.1068\right)\\ =-0.53\end{array}$

Case (iii):

Substitute $0.1\text{s}$ for $t$ in Equation (2).

$\begin{array}{c}5\sin \left(5t-0.15707\right)=5\sin \left(5\left(0.1\text{s}\right)-0.15707\right)\\ =5\sin \left(0.5-0.15707\right)\\ =5\sin \left(0.34293\right)\\ =5\left(0.3362\right)\\ =1.7\end{array}$

Conclusion:

Thus, the value of expression for $t=0\text{s}$ is $\underset{\_}{-0.78}$, for $t=0.01\text{s}$ is $\underset{\_}{-0.53}$, and for $t=0.1\text{s}$ is $\underset{\_}{1.7}$.

To determine

Find the value of following expression.

Answer to Problem 1E

The value of expression for $t=0\text{s}$ is $\underset{\_}{4.0}$, for $t=1\text{s}$ is $\underset{\_}{-1.7}$, and for $t=1.5\text{s}$ is $\underset{\_}{-4.0}$.

Explanation of Solution

Case (i):

Given data:

$4\cos 2t$ (3)

Calculation:

Substitute $0\text{s}$ for $t$ in Equation (3).

$\begin{array}{c}4\cos 2t=4\cos 2\left(0\text{s}\right)\\ =4\cos \left(0\right)\\ =4\left(1\right)\\ =4.0\end{array}$

Case (ii):

Substitute $1\text{s}$ for $t$ in Equation (3).

$\begin{array}{c}4\cos 2t=4\cos 2\left(1\text{s}\right)\\ =4\cos \left(2\right)\\ =4\left(-0.416\right)\\ =-1.7\end{array}$

Case (iii):

Substitute $1.5\text{s}$ for $t$ in Equation (3).

$\begin{array}{c}4\cos 2t=4\cos 2\left(1.5\text{s}\right)\\ =4\cos \left(3\right)\\ =4\left(-0.9899\right)\\ =-4.0\end{array}$

Conclusion:

Thus, the value of expression for $t=0\text{s}$ form is $\underset{\_}{4.0}$, for $t=1\text{s}$ form is $\underset{\_}{-1.7}$, and for $t=1.5\text{s}$ form is $\underset{\_}{-4.0}$.

To determine

Find the value of following expression.

Answer to Problem 1E

The value of expression for $t=0\text{s}$ is $\underset{\_}{3.1}$, for $t=0.01\text{s}$ is $\underset{\_}{3.0}$, and for $t=0.1\text{s}$ is $\underset{\_}{2.1}$.

Explanation of Solution

Case (i):

Given data:

$3.2\cos \left(6t+15°\right)$ (4)

Calculation:

Convert the value in degree to radians.

$\begin{array}{c}3.2\cos \left(6t+15°\right)=3.2\cos \left(6t+15\left(\frac{\pi }{180}\right)\right)\\ =3.2\cos \left(6t+15\left(0.017453\right)\right)\end{array}$

$3.2\cos \left(6t+15°\right)=3.2\cos \left(6t+0.261795\right)$ (5)

Substitute $0\text{s}$ for $t$ in Equation (5).

$\begin{array}{c}3.2\cos \left(6t+15°\right)=3.2\cos \left(6t+15\left(\frac{\pi }{180}\right)\right)\\ =3.2\cos \left(6t+15\left(0.017453\right)\right)\\ =3.2\cos \left(0.261795\right)\\ =3.2\left(0.9659\right)\\ =3.1\end{array}$                                                

Case (ii):

Substitute $0.01\text{s}$ for $t$ in Equation (5).

$\begin{array}{c}3.2\cos \left(6t+0.261795\right)=3.2\cos \left(6\left(0.01\right)+0.261795\right)\\ =3.2\cos \left(0.06+0.261795\right)\\ =3.2\cos \left(0.321795\right)\\ =3.2\left(0.9486\right)\\ =3.0\end{array}$

Case (iii):

Substitute $0.1\text{s}$ for $t$ in Equation (5).

$\begin{array}{c}3.2\cos \left(6t+0.261795\right)=3.2\cos \left(6\left(0.1\right)+0.261795\right)\\ =3.2\cos \left(0.6+0.261795\right)\\ =3.2\cos \left(0.861795\right)\\ =3.2\left(0.6510\right)\\ =2.1\end{array}$

Conclusion:

Thus, the value of expression for $t=0\text{s}$ is $\underset{\_}{3.1}$, for $t=0.01\text{s}$ is $\underset{\_}{3.0}$, and for $t=0.1\text{s}$ is $\underset{\_}{2.1}$.