Chapter 14, Problem 57P

(a)

To determine

The height to which the object eventually rise.

Explanation of Solution

Given:

The mass of the object is $2.0\text{kg}$ .

The length of the object is $5\text{cm}$ .

The amplitude of the object is $8\text{cm}$ .

Formula used:

Write the expression for the maximum speed of the object.

  $v_{\max }=A\omega$ ……. (1)

Here, $v_{\max }$ is the maximum velocity of the object, $A$ is the amplitude of the object and $\omega$ is the angular velocity of the object.

Write the expression for the angular velocity of the object.

  $\omega =\sqrt{\frac{k}{m}}$ …… (2)

Here, $k$ is the spring constant and $m$ is the mass of the object.

Substitute $\sqrt{\frac{k}{m}}$ for $\omega$ in equation (1).

  $v_{\max }=A\sqrt{\frac{k}{m}}$

Solve the above equation for $A$ .

  $A=v_{\max }\sqrt{\frac{m}{k}}$ …… (3)

When object is at equilibrium position, net force on the object is zero.

Force acting in the $y$ direction will be equal that is:

  $k\Delta y-mg=0$

Here, $\Delta y$ is the change in position and $g$ is acceleration due to gravity.

Solve the above equation for $k$ .

  $k=\frac{mg}{\Delta y}$

Substitute $\frac{mg}{\Delta y}$ for $k$ in equation (3).

  $A=v_{\max }\sqrt{\frac{\Delta y}{g}}$

The maximum height of the object is:

  $h=A+5.\text{0}\text{cm}$ …… (4)

Substitute $v_{\max }\sqrt{\frac{\Delta y}{g}}$ for $A$ in equation (4).

  $h=v_{\max }\sqrt{\frac{\Delta y}{g}}+5.\text{0}\text{cm}$ …… (5)

Calculation:

Substitute $0.30\text{m}/{\text{s}}^{\text{2}}$ for $v_{\max }$ , $0.030\text{m}$ for $\Delta y$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (5).

  $\begin{array}{l}h=\text{0}\text{.30}\text{m}/\text{s}\sqrt{\frac{\text{0}\text{.030}\text{m}}{\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}}}\text{+5}\text{.0}\text{cm}\\ h=6.7\text{cm}\end{array}$

Conclusion:

Thus, the maximum height of the object from the floor is $6.7\text{cm}$ .

(b)

To determine

The time taken by the object to reach its maximum height.

Explanation of Solution

Given:

The mass of the object is $2.0\text{kg}$ .

The length of the object is $5\text{cm}$ .

The amplitude of the object is $8\text{cm}$ .

Formula used:

Write the expression for the time period of the oscillator.

  $T=2\pi \sqrt{\frac{m}{k}}$

Substitute $\frac{mg}{\Delta y}$ for $k$ in the above equation.

  $T=2\pi \sqrt{\frac{m}{\frac{mg}{\Delta y}}}$

The time required by the object will be $\frac{3}{4}$ of the time period.

  $t=\frac{3}{4}T$

Substitute $2\pi \sqrt{\frac{m}{\frac{mg}{\Delta y}}}$ for $T$ in above equation.

  $t=\frac{3\pi }{2}\sqrt{\frac{\Delta y}{g}}$ …… (6)

Calculation:

Substitute $0.030\text{m}$ for $\Delta y$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (6).

  $\begin{array}{l}t=\frac{3\pi }{2}\sqrt{\frac{\text{0}\text{.030}\text{m}}{\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}}}\\ t=0.26\text{s}\end{array}$

Conclusion:

Thus, the time the object will take to reach the maximum height is $0.26\text{s}$ .

(c)

To determine

The minimum initial velocity for the object to be upstretched.

Explanation of Solution

Given:

The mass of the object is $2.0\text{kg}$ .

The length of the object is $5\text{cm}$ .

The amplitude of the object is $8\text{cm}$ .

Formula used:

Write the expression for the energy conservation.

  $\Delta K+\Delta U_G+\Delta U_S=0$ …… (7)

Here, $\Delta K$ is the kinetic energy, $\Delta U_G$ is the potential energy and $\Delta U_S$ is the potential energy of the spring.

Substitute $\frac{1}{2}m{v}_i{}^2$ for $\Delta K$ , $-mg\Delta y$ for $\Delta U_G$ and $\frac{1}{2}k\Delta y^2$

  $+\frac{1}{2}k{\left(L-y\right)}^2$ for $\Delta U_S$ in equation (7).

  $\frac{1}{2}m{v}_i{}^2-mg\Delta y+\frac{1}{2}k\Delta y^2-\frac{1}{2}k{\left(L-y\right)}^2=0$ …… (8)

Substitute $\Delta y=L-y$ in equation (8).

  $\begin{array}{l}\frac{1}{2}m{v}_i{}^2-mg\Delta y+\frac{1}{2}k\Delta y^2-\frac{1}{2}{\left(\Delta y\right)}^2=0\\ \frac{1}{2}m{v}_i{}^2-mg\Delta y\end{array}$

Solve the above equation for $v_i$ .

  $v_i=\sqrt{2g\Delta y}$ …… (9)

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $3\text{cm}$ for $\Delta y$ in equation (9).

  $\begin{array}{l}{v}_i=\sqrt{2g\Delta y}{v}_i=\sqrt{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(3\text{cm}\right)}\\ v_i=77\text{cm}/\text{s}\end{array}$

Conclusion:

Thus, the minimum velocity given to the system is $77\text{cm}/\text{s}$ .