Chapter 11, Problem 9P

(a)

To determine

To Find: The velocity at equilibrium point.

Answer to Problem 9P

The velocity at equilibrium point is $2.45\text{ m/s}$ .

Explanation of Solution

Given:

Mass, $m=0.60\text{ kg}$ .

Amplitude, $A=0.13\text{ m}$ .

Frequency, $f=3\text{ Hz}$ .

  $\begin{array}{l}{V}_{\max }=\sqrt{\frac{k}{m}}\times A\\ \sqrt{\frac{k}{m}}=2\pi f\end{array}$

Formula used:

At equilibrium point, the velocity is maximum. It can be obtained by the following formula,

  $\begin{array}{l}{V}_{\max }=\sqrt{\frac{k}{m}}\times A\\ \sqrt{\frac{k}{m}}=2\pi f\end{array}$

Here,

  $V_{\text{max}}$ is maximum velocity,

  $m$ is mass on spring,

  $A$ is the amplitude,

  $k$ is spring constant,

  $f$ is frequency of oscillation.

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{V}_{\max }=\sqrt{\frac{k}{m}}\times A\\ =2\pi f\times A\\ V_{\max }=2\times \pi \times 3\times 0.13=2.45\text{ m/s}\end{array}$

Conclusion:

Thus, the velocity when passes through equilibrium point is $2.45\text{ m/s}$ .

(b)

To determine

To Find:The velocity at some distance from the equilibrium point.

Answer to Problem 9P

The velocity at some distance $0.10\text{ m}$ from the equilibrium point is $1.6\text{ m/s}$ .

Explanation of Solution

Given:

Maximum velocity $V_{\max }=2.45\text{ m/s}$ .

Amplitude $A=0.13\text{ m}$ .

Distance $X=0.10\text{ m}$ .

Formula used:

Velocity at some distance from the equilibrium point is determined by the following formula:

Velocity at different points

  $V=\pm V_{\max }\times \sqrt{1-\frac{X^2}{A^2}}$

Here,

  $V_{\text{max}}$ is maximum velocity,

  $A$ is the amplitude,

  $X$ is the distance at which velocity is to determined.

Calculation:

Substitute the given values,

Velocity at 0.10m from the equilibrium point is,

  $\begin{array}{l}V=\pm 2.45\times \sqrt{1-\frac{{(0.10)}^2}{{(0.13)}^2}}\\ V=1.565\text{ m/s}\end{array}$

Conclusion:

Thus, the velocity at $0.10\text{ m}$ from the equilibrium point is $1.565\text{ m/s}$ .

(c)

To determine

To Find:The total energy.

Answer to Problem 9P

The total energy of the system is $1.8\text{ J}$ .

Explanation of Solution

Given:

Maximum velocity $V_{\max }=2.45\text{ m/s}$ .

Mass $m=0.6\text{ kg}$ .

Formula used:

The total energy of the system is determined by the following formula,

  $E_{\text{total}}=\frac{1}{2}\times m\times V_{\max }{}^2$

Calculation:

Substitute the given values,

  $\begin{array}{l}{E}_{\text{total}}=\frac{1}{2}\times 0.60\times {2.45}^2\\ E_{\text{total}}=1.8\text{ J}\end{array}$

Conclusion:

Thus, the total energy of the system is $1.8\text{ J}$ .

(d)

To determine

To Derive:The equation for motion of mass.

Answer to Problem 9P

The equation describing the motion of mass,where $x$ is maximum at $t=0$ is $x=0.13\cos (6\pi t)$ .

Explanation of Solution

Given:

Amplitude $A=0.13\text{ m}$ .

Frequency $f=3\text{ Hz}$ .

Formula used:

The equation describing the motion of mass, when x is maximum at $t=0$ is shown by the following formula:

  $\begin{array}{l}x=A\cos (\omega t)\\ \omega =2\pi f\end{array}$

Here,

  $A$ is the amplitude.

  $\omega$ is angular frequency of spring.

  $t$ is time.

  $f$ is frequency of oscillation.

Calculation:

Substitute the given values,

  $\begin{array}{l}x=A\cos (\omega t)\\ \omega =2\pi f\\ x=0.13\cos (2\pi \times 3\times t)\\ x=0.13\cos (6\pi t)\end{array}$

Conclusion:

Thus, the equation describing the motion of mass, when $x$ is maximum at $t=0$ is $x=0.13\cos (6\pi t)$ .