Chapter 16, Problem 6PQ

(a)

To determine

The position, velocity and acceleration of each of the simple harmonic oscillator’s at time $t=0\text{s}$

Answer to Problem 6PQ

At $t=0\text{s}$, simple harmonic oscillator’s position is $-3.68\times {10}^{-3}\text{m}$, its velocity is $7.22\times {10}^{-2}\text{m}/\text{s}$ and acceleration is $0.398\text{m}/{\text{s}}^2$ .

Explanation of Solution

Write the expression for the acceleration of the simple harmonic oscillator.

  $a_y\left(t\right)=\left(0.850\text{m}/{\text{s}}^2\right)\cos \left(10.4t-5.20\right)$                                                                    (I)

Here, $a_y\left(t\right)$ is the acceleration of the simple harmonic oscillator.

Write the general equation of acceleration of a simple harmonic oscillator.

  $a_y\left(t\right)=-a_{\text{max}}\cos \left(\omega t+\phi \right)$                                                                                       (II)

Here, $a_{\max }$ is the maximum acceleration, $\omega$ is the angular frequency and $\phi$ is the initial phase.

Write the relation between maximum acceleration and amplitude.

  $a_{\text{max}}=y_{\text{max}}{\omega }^2$                                                                                                        (III)

Here, $y_{\text{max}}$ is the amplitude.

Write the expression for the velocity of simple harmonic oscillator.

  $v_y\left(t\right)=-v_{\text{max}}\sin \left(\omega t+\phi \right)$                                                                                        (IV)

Here, $v_y\left(t\right)$ is the velocity of the harmonic oscillator and $v_{\text{max}}$ is the maximum velocity.

Write the expression for the maximum velocity.

  $v_{\text{max}}=y_{\text{max}}\omega$                                                                                                            (V)

Put equation (V) in (IV) to get relation of $v_y\left(t\right)$.

  $v_y\left(t\right)=-y_{\text{max}}\omega \sin \left(\omega t+\phi \right)$                                                                                    (VI)

Write the expression for the displacement of the simple harmonic oscillator.

  $y\left(t\right)=y_{\text{max}}\cos \left(\omega t+\phi \right)$                                                                                        (VII)

Here, $y\left(t\right)$ is the position of the simple harmonic oscillator.

Conclusion:

Compare equation (I) and (II) to get $\omega$ .

  $\omega =10.4\text{rad}/\text{s}$

Compare equation (I) and (II) to get $a_{\text{max}}$ .

  $a_{\text{max}}=-0.850\text{m}/{\text{s}}^2$

Compare equation (I) and (II) to get $\phi$ .

  $\phi =-5.20$

Substitute $10.4\text{rad}/\text{s}$ for $\omega$ and $0.850\text{m}/{\text{s}}^2$ for $a_{\text{max}}$ in equation (III) to get $y_{\text{max}}$ .

  $\begin{array}{c}-\left(0.850\text{m}/{\text{s}}^2\right)=y_{\text{max}}{\left(10.4\text{rad}/\text{s}\right)}^2\\ y_{\text{max}}=\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\end{array}$

Substitute $\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}$ for $y_{\text{max}}$, $-5.20$ for $\phi$ and $10.4\text{rad}/\text{s}$ for $\omega$ in equation (VI) to get $v_y\left(t\right)$.

  $\begin{array}{c}{v}_y\left(t\right)=-\left(\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\right)\left(10.4\text{rad}/\text{s}\right)\sin \left(\left(10.4\text{rad}/\text{s}\right)t-5.20\right)\\ =\left(\frac{\left(0.850\text{m}/{\text{s}}^2\right)}{\left(10.4\text{rad}/\text{s}\right)}\right)\sin \left(\left(10.4\text{rad}/\text{s}\right)t-5.20\right)\end{array}$                       (VIII)

Substitute $\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}$ for $y_{\text{max}}$, $-5.20$ for $\phi$ and $10.4\text{rad}/\text{s}$ for $\omega$ in equation (VII) to get $y\left(t\right)$.

  $y\left(t\right)=\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\cos \left(\left(10.4\text{rad}/\text{s}\right)t-5.20\right)$                                                  (IX)

Consider $t=0$ .

Substitute $0$ for $t$ in equation (IX) to get $y\left(t\right)$ .

  $\begin{array}{c}y\left(0\right)=\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\cos \left(\left(10.4\text{rad}/\text{s}\right)\left(0\text{s}\right)-5.20\right)\\ =-3.68\text{m}\times {\text{10}}^{-3}\text{m}\end{array}$

Substitute $0$ for $t$ in equation (VIII) to get $v_y\left(t\right)$ .

  $\begin{array}{c}{v}_y\left(0\right)=\left(\frac{\left(0.850\text{m}/{\text{s}}^2\right)}{\left(10.4\text{rad}/\text{s}\right)}\right)\sin \left(\left(10.4\text{rad}/\text{s}\right)\left(0\right)-5.20\right)\\ =7.22\times {10}^{-2}\text{m}/\text{s}\end{array}$

Substitute $0$ for $t$ in equation (I) to get $a_y\left(t\right)$ .

  $\begin{array}{c}{a}_y\left(0\right)=\left(0.850\text{m}/{\text{s}}^2\right)\cos \left(10.4\left(0\text{s}\right)-5.20\right)\\ =0.398\text{m}/{\text{s}}^2\end{array}$

Therefore, at $t=0\text{s}$, simple harmonic oscillators position is $-3.68\times {10}^{-3}\text{m}$, its velocity is $7.22\times {10}^{-2}\text{m}/\text{s}$ and acceleration is $0.398\text{m}/{\text{s}}^2$ .

(b)

To determine

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

Answer to Problem 6PQ

At $t=0.500\text{s}$, simple harmonic oscillator’s position is $-7.86\times {10}^{-3}\text{m}$, its velocity is $0\text{m}/\text{s}$ and its acceleration is $0.850\text{m}/{\text{s}}^2$ .

Explanation of Solution

Use equation (I) to calculate acceleration, equation (VIII) to calculate velocity and equation (IX) to calculate position of simple harmonic oscillator at $t=0.500\text{s}$ .

Conclusion:

Consider $t=0.500\text{s}$ .

Substitute $0.500\text{s}$ for $t$ in equation (IX) to get $y\left(t\right)$ .

  $\begin{array}{c}y\left(0.500\text{s}\right)=\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\cos \left(\left(10.4\text{rad}/\text{s}\right)\left(0.500\text{s}\right)-5.20\right)\\ =-7.86\text{m}\times {\text{10}}^{-3}\text{m}\end{array}$

Substitute $0.500\text{s}$ for $t$ in equation (VIII) to get $v_y\left(t\right)$ .

  $\begin{array}{c}{v}_y\left(0.500\text{s}\right)=\left(\frac{\left(0.850\text{m}/{\text{s}}^2\right)}{\left(10.4\text{rad}/\text{s}\right)}\right)\sin \left(\left(10.4\text{rad}/\text{s}\right)\left(0.500\text{s}\right)-5.20\right)\\ =0\text{m}/\text{s}\end{array}$

Substitute $0.500\text{s}$ for $t$ in equation (I) to get $a_y\left(t\right)$ .

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

Therefore, at $t=0.500\text{s}$, simple harmonic oscillator’s position is $-7.86\times {10}^{-3}\text{m}$, its velocity is $0\text{m}/\text{s}$ and its acceleration is $0.850\text{m}/{\text{s}}^2$ .

(c)

To determine

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

Answer to Problem 6PQ

At $t=2.00\text{s}$, simple harmonic oscillator’s position is $7.81\times {10}^{-3}\text{m}$, its velocity is $8.81\times {10}^{-3}\text{m}/\text{s}$ and its acceleration is $-0.845\text{m}/{\text{s}}^2$ .

Explanation of Solution

Use equation (I) to calculate acceleration, equation (VIII) to calculate velocity and equation (IX) to calculate position of simple harmonic oscillator.

Conclusion:

Consider $t=2.00\text{s}$ .

Substitute $2.00\text{s}$ for $t$ in equation (IX) to get $y\left(t\right)$ .

  $\begin{array}{c}y\left(2.00\text{s}\right)=\frac{-\left(0.850\text{m}/{\text{s}}^2\right)}{{\left(10.4\text{rad}/\text{s}\right)}^2}\cos \left(\left(10.4\text{rad}/\text{s}\right)\left(2.00\text{s}\right)-5.20\right)\\ =7.81\times {\text{10}}^{-3}\text{m}\end{array}$

Substitute $2.00\text{s}$ for $t$ in equation (VIII) to get $v_y\left(t\right)$ .

  $\begin{array}{c}{v}_y\left(2.00\text{s}\right)=\left(\frac{\left(0.850\text{m}/{\text{s}}^2\right)}{\left(10.4\text{rad}/\text{s}\right)}\right)\sin \left(\left(10.4\text{rad}/\text{s}\right)\left(2.00\text{s}\right)-5.20\right)\\ =8.81\times {10}^{-3}\text{m}/\text{s}\end{array}$

Substitute $2.00\text{s}$ for $t$ in equation (I) to get $a_y\left(t\right)$ .

  $\begin{array}{c}{a}_y\left(2.00\text{s}\right)=\left(0.850\text{m}/{\text{s}}^2\right)\cos \left(10.4\left(2.00\text{s}\right)-5.20\right)\\ =-0.845\text{m}/{\text{s}}^2\end{array}$

Therefore, at $t=2.00\text{s}$, simple harmonic oscillator’s position is $7.81\times {10}^{-3}\text{m}$, its velocity is $8.81\times {10}^{-3}\text{m}/\text{s}$ and its acceleration is $-0.845\text{m}/{\text{s}}^2$ .