Chapter 16, Problem 4PQ

(a)

To determine

The position, velocity, and acceleration of simple harmonic oscillator at time $t=0$.

Answer to Problem 4PQ

The position, velocity, and acceleration of simple harmonic oscillator at time $t=0$ is $0.398\text{m}$, $-7.81\text{m}/\text{s}$, and $-43.1\text{m}/{\text{s}}^{\text{2}}$ respectively.

Explanation of Solution

Write the equation to find the position of simple harmonic oscillator.

    $y\left(t\right)=\left(0.850\text{m}\right)\cos \left(10.4t-5.20\right)$                                                                           (I)

Here, $y\left(t\right)$ is the position at time $t$

Differentiate the above expression with respect to $t$ to find the velocity.

  $\begin{array}{c}v\left(t\right)=\frac{d\left(\left(0.850\text{m}\right)\cos \left(10.4t-5.20\right)\right)}{dt}\\ =\left(0.850\text{m}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(10.4t-5.20\right)\end{array}$                                                         (II)

Here, $v\left(t\right)$ is the velocity of particle at time $t$.

Differentiate the above expression with respect to $t$ to find the acceleration.

  $\begin{array}{c}a\left(t\right)=\frac{d\left(\left(0.850\text{m}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(10.4t-5.20\right)\right)}{dt}\\ ={\left(-10.4\text{rad}/\text{s}\right)}^2\left(0.850\text{m}\right)\cos \left(10.4t-5.20\right)\end{array}$                                                 (III)

Conclusion:

Substitute $0\text{s}$ for $t$ in equation (I) to find $y\left(0\right)$.

    $\begin{array}{c}y\left(0\right)=\left(0.850\text{m}\right)\cos \left(10.4\left(0\text{s}\right)-5.20\right)\\ =\left(0.850\text{m}\right)\left(0.468\right)\\ =0.398\text{m}\end{array}$

Substitute $0\text{s}$ for $t$ in equation (II) to find $v\left(0\right)$.

    $\begin{array}{c}v\left(0\right)=\left(0.850\text{m}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(10.4\left(0\text{s}\right)-5.20\right)\\ =\left(8.84\text{m}/\text{s}\right)\left(-0.883\right)\\ =-7.81\text{m}/\text{s}\end{array}$

Substitute $0\text{s}$ for $t$ in equation (III) to find $a\left(0\right)$.

    $\begin{array}{c}a\left(0\right)={\left(-10.4\text{rad}/\text{s}\right)}^2\left(0.850\text{m}\right)\cos \left(10.4\left(0\text{s}\right)-5.20\right)\\ =-91.9\text{m}/{\text{s}}^{\text{2}}\left(0.469\right)\\ =-43.1\text{m}/{\text{s}}^{\text{2}}\end{array}$

Therefore, The position, velocity, and acceleration of simple harmonic oscillator at time $t=0$ is $0.398\text{m}$, $-7.81\text{m}/\text{s}$, and $-43.1\text{m}/{\text{s}}^{\text{2}}$ respectively.

(b)

To determine

The position, velocity, and acceleration of simple harmonic oscillator at time $t=0.500\text{s}$.

Answer to Problem 4PQ

The position, velocity, and acceleration of simple harmonic oscillator at time $t=0.500\text{s}$ is $0.850\text{m}$, $0\text{m}/\text{s}$, and $-91.9\text{m}/{\text{s}}^{\text{2}}$ respectively.

Explanation of Solution

Substitute $0.500\text{s}$ for $t$ in equation (I) to find $y\left(0\right)$.

    $\begin{array}{c}y\left(0\right)=\left(0.850\text{m}\right)\cos \left(10.4\left(0.500\text{s}\right)-5.20\right)\\ =\left(0.850\text{m}\right)\left(1\right)\\ =0.850\text{m}\end{array}$

Substitute $0.500\text{s}$ for $t$ in equation (II) to find $v\left(0\right)$.

    $\begin{array}{c}v\left(0\right)=\left(0.850\text{m}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(10.4\left(0.500\text{s}\right)-5.20\right)\\ =\left(8.84\text{m}/\text{s}\right)\left(0\right)\\ =0\text{m}/\text{s}\end{array}$

Substitute $0.500\text{s}$ for $t$ in equation (III) to find $a\left(0\right)$.

    $\begin{array}{c}a\left(0\right)={\left(-10.4\text{rad}/\text{s}\right)}^2\left(0.850\text{m}\right)\cos \left(10.4\left(0.500\text{s}\right)-5.20\right)\\ =-91.9\text{m}/{\text{s}}^{\text{2}}\left(1\right)\\ =-91.9\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the position, velocity, and acceleration of simple harmonic oscillator at time $t=0.500\text{s}$ is $0.850\text{m}$, $0\text{m}/\text{s}$, and $-91.9\text{m}/{\text{s}}^{\text{2}}$ respectively.

(c)

To determine

The position, velocity, and acceleration of simple harmonic oscillator at time $t=2.00\text{s}$.

Answer to Problem 4PQ

The position, velocity, and acceleration of simple harmonic oscillator at time $t=2.00\text{s}$ is $-0.845\text{m}$, $0.953\text{m}/\text{s}$, and $91.4\text{m}/{\text{s}}^{\text{2}}$ respectively.

Explanation of Solution

Substitute $2.00\text{s}$ for $t$ in equation (I) to find $y\left(0\right)$.

    $\begin{array}{c}y\left(0\right)=\left(0.850\text{m}\right)\cos \left(10.4\left(2.00\text{s}\right)-5.20\right)\\ =\left(0.850\text{m}\right)\left(-0.994\right)\\ =-0.845\text{m}\end{array}$

Substitute $2.00\text{s}$ for $t$ in equation (II) to find $v\left(0\right)$.

    $\begin{array}{c}v\left(0\right)=\left(0.850\text{m}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(10.4\left(2.00\text{s}\right)-5.20\right)\\ =\left(8.84\text{m}/\text{s}\right)\left(0.107\right)\\ =0.953\text{m}/\text{s}\end{array}$

Substitute $2.00\text{s}$ for $t$ in equation (III) to find $a\left(0\right)$.

    $\begin{array}{c}a\left(0\right)={\left(-10.4\text{rad}/\text{s}\right)}^2\left(0.850\text{m}\right)\cos \left(10.4\left(2.00\text{s}\right)-5.20\right)\\ =91.9\text{m}/{\text{s}}^{\text{2}}\left(0.994\right)\\ =91.4\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the position, velocity, and acceleration of simple harmonic oscillator at time $t=2.00\text{s}$ is $-0.845\text{m}$, $0.953\text{m}/\text{s}$, and $91.4\text{m}/{\text{s}}^{\text{2}}$ respectively.