Chapter 14, Problem 52P

(a)

To determine

The mass of the object.

Explanation of Solution

Given:

The force constant is $1800\text{N}/\text{m}$ .

The amplitude of the object is $2.50\text{cm}$ .

The frequency of oscillation is $5.50\text{Hz}$.

Formula used:

Write the expression for the angular frequency of the object.

  $\omega =\sqrt{\frac{k}{m}}$

Here, $\omega$ is the angular frequency of oscillation, $k$ is the force constant of the spring and $m$ is the mass of the object.

Simplify the above equation for $m$ .

  $m=\frac{k}{{\omega }^2}$ …… (1)

Write the expression for angular frequency in terms of frequency.

  $\omega =2\pi f$

Here, $f$ is the frequency of oscillation.

Substitute $2\pi f$ for $\omega$ in equation (1).

  $m=\frac{k}{4{\pi }^2{f}^2}$ …… (2)

Calculation:

Substitute $5.50\text{Hz}$ for $f$ and $1800\text{N}/\text{m}$ for $k$ in equation (2).

  $\begin{array}{l}m=\frac{1800\text{N}/\text{m}}{4{\pi }^2{\left(5.50\text{Hz}\right)}^2}\\ m=1.507\text{kg}\end{array}$

Conclusion:

Thus, the mass of the object is $1.507\text{kg}$.

(b)

To determine

The amount of spring stretched from its upstretched length when the object is in equilibrium.

Explanation of Solution

Given:

The force constant is $1800\text{N}/\text{m}$ .

The amplitude of the object is $2.50\text{cm}$ .

The frequency of oscillation is $5.50\text{Hz}$.

Formula used:

Assume $\Delta x$ to be the amount the spring is stretched from its original length when the object is in equilibrium.

The force at the equilibrium position $\sum F$ is zero.

Write the expression for the net force on the object at equilibrium position.

  $k\Delta x-mg=0$ …… (3)

Here, $\Delta x$ is the amount the spring is stretched from the equilibrium position, $g$ is the acceleration due to gravity.

Rearrange equation (3) in terms of $\Delta x$ .

  $\Delta x=\frac{mg}{k}$ …… (4)

Calculation:

Substitute $1.507\text{kg}$ for $m$ , $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1800\text{N}/\text{m}$ for $k$ in equation (4).

  $\begin{array}{l}\Delta x=\frac{1.507kg\cdot 9.81m/s^2}{1800\text{N}/\text{m}}\\ \Delta x=8.21\text{mm}\end{array}$

Conclusion:

Thus, theamount the spring is stretched from its original length when the object is in equilibrium is $8.21\text{mm}$ .

(c)

To determine

The expression for the position, velocity and acceleration of the object.

Explanation of Solution

Given:

The force constant is $1800\text{N}/\text{m}$ .

The amplitude of the object is $2.50\text{cm}$ .

The frequency of oscillation is $5.50\text{Hz}$.

Formula used:

Write the expression for the position of object as the function of time.

  $x=A\cos \left(\omega t+\delta \right)$ …… (5)

Here, $x$ is the position of the particle $\omega$ is the angular frequency of the particle and $\delta$ is the phase constant.

Differentiate the position of the particle to determine velocity.

  $v_x=-A\omega \left(\sin \left(\omega t+\delta \right)\right)$ …… (6)

Here, $v_x$ is the velocity of the particle.

Divide equation (6) and (5) to obtain $\delta$

  $\delta ={\tan }^{-1}\left(\frac{-v_x}{\omega x}\right)$ …… (7)

Substitute $2.50\text{cm}$ for $x$ and $0$ for $v_x$ in equation (7).

  $\begin{array}{l}\delta ={\tan }^{-1}\left(\frac{-0}{\omega \times 2.50\text{cm}}\right)\\ \delta =\pi \end{array}$

Write the expression for the frequency of oscillation.

  $\omega =\sqrt{\frac{k}{m}}$ …… (8)

Substitute $1800\text{N}/\text{m}$ for $k$ and $1.507\text{kg}$ for $m$ in equation (8).

  $\begin{array}{l}\omega =\sqrt{\frac{\text{1800}\text{N}/\text{m}}{1.507\text{kg}}}\\ \omega =34.56\text{rad}/\text{s}\end{array}$

Calculation:

Substitute $34.56\text{rad}/\text{s}$ for $\omega$ , $2.50\text{cm}$ for $A$ and $\pi$ for $\delta$ in equation (5).

  $\begin{array}{l}x=2.50\text{cm}\left(\cos \left(34.56\text{rad}/\text{s}\text{+}\pi \right)\right)\\ x=-2.50\text{cm}\left(\cos \left(34.56\text{rad}/\text{s}\right)t\right)\end{array}$

Obtain the velocity of the object.

Substitute $34.56\text{rad}/\text{s}$ for $\omega$ , $2.50\text{cm}$ for $A$ and $\pi$ for $\delta$ in equation (6).

  $\begin{array}{l}{v}_x=-2.50\text{cm}\cdot \text{34}\text{.56}\text{rad}/\text{s}\left(\sin \left(\text{34}\text{.56}\text{rad}/\text{s}.t+\pi \right)\right)\\ v_x=86.4\text{cm}/\text{s}\left(\sin \left(\text{34}\text{.56}\text{rad}/\text{s}.t\right)\right)\end{array}$

Differentiate equation (6) to obtain acceleration.

  $a_x=-A{\omega }^2\sin \left(\omega t+\delta \right)$ …… (9)

Here, $a_x$ is the acceleration of the particle.

  $\begin{array}{l}{a}_x=-2.50\text{cm}\cdot {\left(\text{34}\text{.56}\text{rad}/\text{s}\right)}^2\sin \left(\left(\text{34}\text{.56}\text{rad}/\text{s}\right)t+\pi \right)\\ a_x=29.9\text{m}/{\text{s}}^{\text{2}}\cos \left(\left(\text{34}\text{.56}\text{rad}/\text{s}\right)t\right)\end{array}$

Conclusion:

Thus, the position of the object is $-2.50\text{cm}\left(\cos \left(34.56\text{rad}/\text{s}\right)t\right)$ , the velocity of the object is $86.4\text{cm}/\text{s}\left(\sin \left(\text{34}\text{.56}\text{rad}/\text{s}.t\right)\right)$ and the acceleration of the object is $29.9\text{m}/{\text{s}}^{\text{2}}\cos \left(\left(\text{34}\text{.56}\text{rad}/\text{s}\right)t\right)$ .