Chapter 12, Problem 11P

(a)

To determine

The position of the particle at the end of $t=4.50\text{s}$.

Answer to Problem 11P

The position of the particle at the end of $t=4.50\text{s}$ is $-2.34\text{ m}$.

Explanation of Solution

Given information:

The initial position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The formula for the position of the particle is,

$x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

$x_0$ is the initial position of the particle.

$v_0$ is the initial velocity of the particle.

$t$ is the time.

$a$ is the acceleration of the particle.

Substitute $0.270\text{m}$ for $x_0$, $0.140\text{m}/\text{s}$ for $v_0$, $4.50\text{s}$ for $t$ and $-0.320\text{m}/{\text{s}}^2$ for $a$ in above equation to find $x$.

$\begin{array}{c}x=\left(0.270\text{m}\right)+\left(0.140\text{m}/\text{s}\right)\left(4.50\text{s}\right)+\frac{1}{2}\left(-0.320\text{m}/{\text{s}}^2\right){\left(4.50\text{s}\right)}^2\\ =-2.34\text{ m}\end{array}$

Conclusion:

Therefore, the position of the particle at the end of $t=4.50\text{s}$ is $-2.34\text{ m}$.

(b)

To determine

The velocity of the particle at the end of $t=4.50\text{s}$.

Answer to Problem 11P

The velocity of the particle at the end of $t=4.50\text{s}$ is $-1.30\text{m}/\text{s}$.

Explanation of Solution

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The formula for the velocity of the particle is,

$v=v_0+at$

Substitute $0.140\text{m}/\text{s}$ for $v_0$, $4.50\text{s}$ for $t$ and $-0.320\text{m}/{\text{s}}^2$ for $a$ in above equation to find $v$.

$\begin{array}{c}v=\left(0.140\text{m}/\text{s}\right)+\left(-0.320\text{m}/{\text{s}}^2\right)\left(4.50\text{s}\right)\\ =-1.30\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the particle at the end of $t=4.50\text{s}$ is $-1.30\text{m}/\text{s}$.

(c)

To determine

The position of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$.

Answer to Problem 11P

The position of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$ is $-0.078\text{m}$.

Explanation of Solution

Section 1:

To determine: The angular frequency of the particle.

Answer: The angular frequency of the particle is $1.08\text{rad}/\text{s}$.

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The formula for the acceleration of the particle is,

$a=-{\omega }^2{x}_0$

$\omega$ is the angular frequency.

Substitute $0.270\text{m}$ for $x_0$ and $-0.320\text{m}/{\text{s}}^2$ for $a$ in above equation to find $\omega$.

$\begin{array}{c}\left(-0.320\text{m}/{\text{s}}^2\right)=-{\omega }^2\left(0.270\text{m}\right)\\ \omega =\sqrt{\frac{0.320\text{m}/{\text{s}}^2}{0.270\text{m}}}\\ =1.08\text{rad}/\text{s}\end{array}$

Section 2:

To determine: The amplitude of the motion.

Answer: The amplitude of the motion is $0.29\text{m}$.

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The general form of position of the particle is,

$x=A\cos \left(\omega t+\varphi \right)$

$A$ is the amplitude.

$\varphi$ is the phase constant.

At the time $t=0$, the position is $x=0.270\text{m}$.

Substitute $0$ for $t$ and $0.270\text{m}$ for $x$ in above equation.

$\begin{array}{c}0.270\text{m}=A\cos \left(\omega \left(0\right)+\varphi \right)\\ =A\cos \left(\varphi \right)\\ \cos \left(\varphi \right)=\left(\frac{0.270\text{m}}{A}\right)\end{array}$ (I)

The general form of velocity of the particle is,

$v=-\omega A\sin \left(\omega t+\varphi \right)$

Substitute $0$ for $t$, $1.08\text{rad}/\text{s}$ for $\omega$ and $0.140\text{m}/\text{s}$ for $v_0$ in above equation.

$\begin{array}{c}\left(0.140\text{m}/\text{s}\right)=-\left(1.08\text{rad}/\text{s}\right)A\sin \left(\omega \left(0\right)+\varphi \right)\\ =-\left(1.08\text{rad}/\text{s}\right)A\sin \left(\varphi \right)\\ \sin \left(\varphi \right)=-\frac{\left(0.140\text{m}/\text{s}\right)}{\left(1.08\text{rad}/\text{s}\right)A}\\ =-\frac{\left(0.12\text{m}\right)}{A}\end{array}$ (II)

Solve the equation (I) and equation (II) to obtain value of $A$.

$\begin{array}{c}{\sin }^2\left(\varphi \right)+{\cos }^2\left(\varphi \right)={\left(-\frac{0.12\text{m}}{A}\right)}^2+{\left(\frac{0.270\text{m}}{A}\right)}^2\\ 1=\frac{0.014{\text{m}}^2}{A^2}+\frac{0.072{\text{m}}^2}{A^2}\\ A=\sqrt{0.014{\text{m}}^2+0.072{\text{m}}^2}\\ =0.29\text{m}\end{array}$

Section 3:

To determine: The phase constant of the motion.

Answer: The phase constant of the motion is $-24.2°$.

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

Substitute $0.29\text{m}$ for $A$ in equation (II) to find $\varphi$.

$\begin{array}{c}\sin \left(\varphi \right)=\left(-\frac{0.12\text{m}}{0.29\text{m}}\right)\\ \varphi ={\sin }^{-1}\left(-\frac{0.12\text{m}}{0.29\text{m}}\right)\\ =-24.2°\end{array}$

Section 4:

To determine: The position of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$.

Answer: The position of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$ is $-0.078\text{m}$.

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The formula for the position of the particle is,

$x=A\cos \left(\omega t+\varphi \right)$

Substitute $0.29\text{m}$ for $A$, $1.08\text{rad}/\text{s}$ for $\omega$, $4.50\text{s}$ for $t$ and $-24.2°$ for $\varphi$ in above equation to find $x$.

$\begin{array}{c}x=\left(0.29\text{m}\right)\cos \left(\left(1.08\text{rad}/\text{s}\right)\left(4.50\text{s}\right)-24.2°\right)\\ =\left(0.29\text{m}\right)\cos \left(4.86\text{rad}\left(\frac{180°}{\text{π}}\right)-24.2°\right)\\ =\left(0.29\text{m}\right)\cos \left(278.5°-24.2°\right)\\ =-0.078\text{m}\end{array}$

Conclusion:

Therefore, the position of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$ is $-0.078\text{m}$.

(d)

To determine

The velocity of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$.

Answer to Problem 11P

The velocity of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$ is $0.30\text{m}/\text{s}$.

Explanation of Solution

Given information:

The position of the particle is $0.270\text{m}$, the velocity of the particle is $0.140\text{m}/\text{s}$, the acceleration of the particle is $-0.320\text{m}/{\text{s}}^2$ and the time is $4.50\text{s}$.

The general form of velocity of the particle is,

$v=-\omega A\sin \left(\omega t+\varphi \right)$

Substitute $0.29\text{m}$ for $A$, $1.08\text{rad}/\text{s}$ for $\omega$, $4.50\text{s}$ for $t$ and $-24.2°$ for $\varphi$ in above equation to find $v$.

$\begin{array}{c}v=-\left(1.08\text{rad}/\text{s}\right)\left(0.29\text{m}\right)\sin \left(\left(1.08\text{rad}/\text{s}\right)\left(4.50\text{s}\right)-24.2°\right)\\ =-\left(1.08\text{rad}/\text{s}\right)\left(0.29\text{m}\right)\sin \left(4.86\text{rad}\left(\frac{180°}{\text{π}}\right)-24.2°\right)\\ =-\left(1.08\text{rad}/\text{s}\right)\left(0.29\text{m}\right)\sin \left(278.5°-24.2°\right)\\ =0.30\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the particle in simple harmonic motion for $t=4.50\text{s}$ and $x=0$ is $0.30\text{m}/\text{s}$.