Chapter 16, Problem 25PQ

A spring of mass m_s and spring constant k is attached to an object of mass M and set into simple harmonic motion on a frictionless, horizontal table. All portions of the spring are assumed to oscillate in phase, and the velocity of each segment dx of the spring with mass dm can be assumed to be proportional to the distance x of that segment from point A in Figure P16.25. a. What is the kinetic energy of the system at the instant the object is moving with speed v? b. What is the frequency of oscillation of the system?

FIGURE P16.25

To determine

The kinetic energy of the system at the instant the object is moving with speed $v$ .

Answer to Problem 25PQ

The kinetic energy of the system at the instant the object is moving with speed $v$ is $\frac{1}{2}\left(M+\frac{m}{3}\right)v^2$ .

Explanation of Solution

It is given that velocity of each segment $dx$ of the spring with mass $dm$ can be assumed to be proportional to the distance $x$ of that segment from point $A$ in figure P16.25.

Write the expression for the kinetic energy of segment of mass $dm$ .

  $dK=\frac{1}{2}\left(dm\right)v_x^2$                                                                                                         (I)

Here, $dK$ is the kinetic energy of the segment of mass $dm$ and length $dx$ and $v_x$ is the velocity of segment at distance $x$ from $A$ .

The velocity of segment is a function of distance from $A$. Thus, velocity of segment at fixed end of the spring is $0$ and that at end of spring is equal to that of speed of mass.

This indicates that velocity of each segment varies linearly from $0$ to speed of mass.

Write the expression for the velocity of segment at distance $x$ from $A$ .

  $v_x=\frac{x}{l}v$                                                                                                                    (II)

Here, $l$ is the length of spring, $x$ is the distance from $x$ and $v$ is the speed of mass.

Write the expression for mass of segment.

  $dm=\frac{m}{l}dx$                                                                                                              (III)

Here, $m$ is the total mass of spring.

The kinetic energy of the system is the sum of kinetic energy of mass and total kinetic energy of the spring.

Write the expression for the total kinetic energy of system.

  $K=K_M+K_{\text{spring}}$                                                                                                      (IV)

Here, $K$ is the kinetic energy of the system, $K_M$ is the kinetic energy of object of mass $M$ and $K_{\text{spring}}$ is the kinetic energy of spring.

Write the expression for the kinetic energy of object.

  $K_M=\frac{1}{2}M{v}^2$                                                                                                            (V)

Here, $M$ is the mass object and $v$ speed of object.

Write the integral equation to find total kinetic energy of spring.

  $K_{\text{spring}}=\frac{1}{2}{\displaystyle \underset{0}{\overset{l}{\int }}{v}_x^{2}dm}$                                                                                                     (VI)

Substitute (II) and (III) in (VI) to get $K_{\text{spring}}$ .

  $K_{\text{spring}}=\frac{1}{2}{\displaystyle \underset{0}{\overset{l}{\int }}{\left(\frac{x}{l}v\right)}^2}\frac{m}{l}dx$

Integrate above equation to get $K_{\text{spring}}$ .

  $\begin{array}{c}{K}_{\text{spring}}=\frac{1}{2}{\displaystyle \underset{0}{\overset{l}{\int }}{\left(\frac{x}{l}v\right)}^2\frac{m}{l}dx}\\ =\frac{1}{2}\frac{v^{2}m}{l^3}{\displaystyle \underset{0}{\overset{l}{\int }}{x}^{2}dx}\\ =\frac{1}{2}\frac{v^{2}m}{l^3}{\left[\frac{x^3}{3}\right]}_0^l\end{array}$

Apply upper limit and lower limit to get $K_{\text{spring}}$ .

  $\begin{array}{c}{K}_{\text{spring}}=\frac{1}{2}\frac{v^{2}m}{l^2}{\left[\frac{x^3}{3}\right]}_0^l\\ =\frac{1}{2}\frac{v^{2}m}{l^3}\left[\frac{l^3}{3}\right]\\ =\frac{1}{2}\frac{m{v}^2}{3}\end{array}$

Substitute $\frac{1}{2}\frac{m{v}^2}{3}$ for $K_{\text{spring}}$ and $\frac{1}{2}M{v}^2$ for $K_M$ in (IV) to get $K$ .

  $\begin{array}{c}K=\frac{1}{2}M{v}^2+\frac{1}{2}\frac{m{v}^2}{3}\\ =\frac{1}{2}\left(M+\frac{m}{3}\right)v^2\end{array}$                                                                                       (VII)

Conclusion:

Therefore, the kinetic energy of the system at the instant the object is moving with speed $v$ is $\frac{1}{2}\left(M+\frac{m}{3}\right)v^2$ .

To determine

The frequency of oscillation of the system.

Answer to Problem 25PQ

The frequency of oscillation of the system is $\frac{1}{2\pi }\sqrt{\frac{k}{M+\frac{m}{3}}}$ .

Explanation of Solution

Write the expression for the frequency of oscillation.

  $f=\frac{\omega }{2\pi }$                                                                                                                 (VIII)

Here, $f$ is the frequency of oscillation of mass-spring system and $\omega$ is the angular frequency of the system.

Write the expression for $\omega$ .

  $\omega =\sqrt{\frac{k}{m_{\text{eff}}}}$

Here, $k$ is the spring constant and $m_{\text{eff}}$ is the effective mass of the system.

Substitute $\sqrt{\frac{k}{m_{\text{eff}}}}$ for $\omega$ in (VIII) to get $f$ .

  $f=\frac{1}{2\pi }\sqrt{\frac{k}{m_{\text{eff}}}}$                                                                                                         (IX)

Since mass is not located at end of the spring, the system must be considered to have spring constant and effective mass. Using total mass frequency of oscillation cannot be obtained.

Write expression for kinetic energy of system.

  $K=\frac{1}{2}{m}_{\text{eff}}{v}^2$

The kinetic energy is also equal that calculated in part (a).

Equate above equation with (VII) to get $m_{\text{eff}}$.

  $\begin{array}{c}\frac{1}{2}{m}_{\text{eff}}{v}^2=\frac{1}{2}\left(M+\frac{m}{3}\right)v^2\\ m_{\text{eff}}=M+\frac{m}{3}\end{array}$

Conclusion:

Substitute $M+\frac{m}{3}$ for $m_{\text{eff}}$ in (IX) to get $f$ .

  $f=\frac{1}{2\pi }\sqrt{\frac{k}{M+\frac{m}{3}}}$

Therefore, the frequency of oscillation of the system is $\frac{1}{2\pi }\sqrt{\frac{k}{M+\frac{m}{3}}}$