Chapter 12, Problem 9P

(a)

To determine

The expression for the position of the particle as a function of time.

Answer to Problem 9P

The expression for the position of the particle as a function of time is $\left(2.00\text{cm}\right)\sin \left(\text{3π}t\right)$.

Explanation of Solution

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The expression for the position of the particle for simple harmonic motion is,

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

$A$ is the amplitude.

$\omega$ is the angular frequency.

$t$ is the time.

$\varphi$ is the phase constant.

The formula to calculate angular frequency is,

$\omega =\text{2π}f$

$f$ is the frequency.

Substitute $1.50\text{Hz}$ for $f$ in above equation.

$\begin{array}{c}\omega =\text{2π}\left(1.50\text{Hz}\right)\\ =3\text{π}\end{array}$

For $t=0$, the position $x=0$ so the phase constant is $0°$ at this instant of time.

Substitute $2.00\text{cm}$ for $A$, $0°$ for $\varphi$ and $3\text{π}$ for $\omega$ in above equation (I) to find $x$.

$\begin{array}{c}x=\left(2.00\text{cm}\right)\sin \left(\text{3π}t-0\right)\\ =\left(2.00\text{cm}\right)\sin \left(\text{3π}t\right)\end{array}$ (II)

Conclusion:

Therefore, the expression for the position of the particle as a function of time is $\left(2.00\text{cm}\right)\sin \left(\text{3π}t\right)$.

(b)

To determine

The maximum speed of the particle.

Answer to Problem 9P

The maximum speed of the particle is $18.8\text{cm}/\text{s}$.

Explanation of Solution

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The formula to calculate maximum velocity is,

$v=A\omega$

Substitute $2.00\text{cm}$ for $A$ and $\text{3π}$ for $\omega$ in above equation to find $v$.

$\begin{array}{c}v=\left(2.00\text{cm}\right)\text{3π}\\ =18.8\text{cm}/\text{s}\end{array}$

Conclusion:

Therefore, the maximum speed of the particle is $18.8\text{cm}/\text{s}$.

(c)

To determine

The earliest time at which the particle has $18.8\text{cm}/\text{s}$ speed.

Answer to Problem 9P

The earliest time at which the particle has $18.8\text{cm}/\text{s}$ speed is $0.33\text{s}$.

Explanation of Solution

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The expression for velocity is,

$v=\frac{dx}{dt}$

Substitute $\left(2.00\text{cm}\right)\sin \left(\text{3π}t\right)$ for $x$ and differentiate with respect to time.

$\begin{array}{c}v=\frac{d\left(\left(2.00\text{cm}\right)\sin \left(\text{3π}t\right)\right)}{dt}\\ =\left(\text{3π}\right)\left(2.00\text{cm}\right)\cos \left(\text{3π}t\right)\\ =\left(18.8\text{cm}/\text{s}\right)\cos \left(\text{3π}t\right)\end{array}$

Substitute $18.8\text{cm}/\text{s}$ for $v$ in above equation to find $t$.

$\begin{array}{c}\left(18.8\text{cm}/\text{s}\right)=\left(18.8\text{cm}/\text{s}\right)\cos \left(\text{3π}t\right)\\ \cos \left(\text{3π}t\right)=1\\ \text{3π}t=\text{π}\\ t=0.33\text{s}\end{array}$

Conclusion:

Therefore, the earliest time at which the particle has $18.8\text{cm}/\text{s}$ speed is $0.33\text{s}$.

(d)

To determine

The maximum positive acceleration of the particle.

Answer to Problem 9P

The maximum positive acceleration of the particle is $177.6\text{cm}/{\text{s}}^2$.

Explanation of Solution

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The expression for acceleration is,

$a=A{\omega }^2$

Substitute $2.00\text{cm}$ for $A$ and $\text{3π}$ for $\omega$ in above equation to find $a$.

$\begin{array}{c}a=\left(2.00\text{cm}\right){\left(\text{3π}\right)}^2\\ =177.6\text{cm}/{\text{s}}^2\end{array}$

Conclusion:

Therefore, the maximum positive acceleration of the particle is $177.6\text{cm}/{\text{s}}^2$.

(e)

To determine

The earliest time at which the particle has $177.6\text{cm}/{\text{s}}^2$ acceleration.

Answer to Problem 9P

The earliest time at which the particle has $177.6\text{cm}/{\text{s}}^2$ acceleration is $0.5\text{s}$.

Explanation of Solution

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The expression for velocity is,

$a=\frac{dv}{dt}$

Substitute $\left(18.8\text{cm}/\text{s}\right)\cos \left(\text{3π}t\right)$ for $v$ and differentiate with respect to time.

  $\begin{array}{c}a=\frac{d\left(\left(18.8\text{cm}/\text{s}\right)\cos \left(\text{3π}t\right)\right)}{dt}\\ =-\text{3π}\left(18.8\text{cm}/\text{s}\right)\sin \left(\text{3π}t\right)\\ =-\left(177.6\text{cm}/{\text{s}}^2\right)\sin \left(\text{3π}t\right)\end{array}$

Substitute $177.6\text{cm}/{\text{s}}^2$ for $a$ in above equation to find $t$.

$\begin{array}{c}\left(177.6\text{cm}/{\text{s}}^2\right)=-\left(177.6\text{cm}/{\text{s}}^2\right)\sin \left(\text{3π}t\right)\\ \sin \left(\text{3π}t\right)=-1\\ \text{3π}t=\frac{\text{3π}}{2}\\ t=0.5\text{s}\end{array}$

Conclusion:

Therefore, the earliest time at which the particle has $177.6\text{cm}/{\text{s}}^2$ acceleration is $0.5\text{s}$.

(f)

To determine

The total distance traveled by the particle between $t=0$ to $t=1.00\text{s}$.

Answer to Problem 9P

The total distance traveled by the particle between $t=0$ to $t=1.00\text{s}$ is $12.0\text{cm}$.

Explanation of Solution

Section 1;

To determine: The time period of the particle.

Answer: The time period of the particle is $0.66\text{s}$.

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The time period of the particle is,

$T=\frac{1}{f}$

Substitute $1.5\text{Hz}$ for $f$ in above equation to find $T$.

$\begin{array}{c}T=\frac{1}{1.5\text{Hz}}\\ =0.66\text{s}\end{array}$

Section 2;

To determine: The number of time period of the particle.

Answer: The number of time period of the particle is $1.5$.

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

The number of time periods is calculated as,

$\begin{array}{c}n=\frac{1.00\text{s}}{0.66\text{s}}\\ =1.5\end{array}$

This number of the periods shows that it completes one and half cycle approximately.

Section 3;

To determine: The total distance traveled by the particle between $t=0$ to $t=1.00\text{s}$.

Answer: The total distance traveled by the particle between $t=0$ to $t=1.00\text{s}$ is $12\text{cm}$.

Given information:

The amplitude of the motion of the particle is $2.00\text{cm}$ and the frequency of the motion of the particle is $1.50\text{Hz}$.

For one and half cycle the total distance is given as,

$\begin{array}{c}x=4A+2A\\ =6A\end{array}$

Substitute $2.00\text{cm}$ for $A$ in above equation to find $x$.

$\begin{array}{c}x=6\left(2.00\text{cm}\right)\\ =12\text{cm}\end{array}$

Conclusion:

Therefore, the total distance traveled by the particle between $t=0$ to $t=1.00\text{s}$ is $12\text{cm}$.