Chapter 11, Problem 98P

(a)

To determine

The oscillation frequency of the pendulum.

Answer to Problem 98P

The oscillation frequency is $\text{0}\text{.133}\text{Hz}$.

Explanation of Solution

Mass of object is $15.0\text{kg}$  and length of thin wire is $14.0\text{m}$.

Write the equation for the frequency of Foucault pendulum.

$f=\frac{1}{2\pi }\sqrt{\frac{g}{L}}$

Here, the frequency of pendulum is $f$, $g$ is the gravitational acceleration, and $L$ is the length Foucault pendulum.

Conclusion:

Substitute $3.14$ for $\pi$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $14.0\text{m}$ for $L$ in the above equation to find $f$.

$\begin{array}{c}f=\frac{1}{2\left(3.14\right)}\sqrt{\frac{9.80\text{m}/{\text{s}}^{\text{2}}}{14.0\text{m}}}\\ =\left(0.159\right)\left(0.837{\text{s}}^{-1}\right)\\ =0.133\text{Hz}\end{array}$

Therefore, the oscillation frequency is $\text{0}\text{.133}\text{Hz}$.

(b)

To determine

The maximum speed of the pendulum.

Answer to Problem 98P

The maximum speed is $\text{1}\text{.24}\text{m}/\text{s}$.

Explanation of Solution

Mass of object is $15.0\text{kg}$, length of thin wire is $14.0\text{m}$, and the maximum oscillation angle of pendulum is $6.10°$

Write the relation between $k$ and speed of wave.

$v=\omega A$ (I)

Here, $v$ is the maximum speed of pendulum,$\omega$ is the angular frequency, and $A$ is the amplitude of wave.

Write the equation for $\omega$.

$\omega =\sqrt{\frac{g}{L}}$ (II)

Refer the diagram shown below. Write the relation between $A$ and $L$.

$A=L\sin \theta$ (III)

Here, $\theta$ is the maximum angle of oscillation.

Rewrite equation (I) by substituting equations (II) and (III).

$\begin{array}{c}v=\sqrt{\frac{g}{L}}L\sin \theta \\ =\sqrt{gL}\sin \theta \end{array}$

Conclusion:

Substitute $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, $14.0\text{m}$ for $L$ , and $6.10°$ for $\theta$ in the above equation to find $v$.

$\begin{array}{c}v=\sqrt{\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\left(14.0\text{m}\right)}\left(\sin 6.10°\right)\\ =\left(11.71\text{m}/\text{s}\right)\left(0.106\right)\\ =1.24\text{m}/\text{s}\end{array}$

Therefore, the maximum speed is $\text{1}\text{.24}\text{m}/\text{s}$.

(c)

To determine

The maximum tension in the wire.

Answer to Problem 98P

The maximum tension is $149\text{N}$.

Explanation of Solution

Mass of object is $15.0\text{kg}$, length of thin wire is $14.0\text{m}$, and the maximum oscillation angle of pendulum is $6.10°$

The free body diagram of object is shown below.

Write the Newton’s second law to find the net force.

$F-mg=m\left(\frac{v^2}{L}\right)$

Here, $F$ is the maximum tension and $m$ is the mass of object.

Rewrite the above equation in terms of $F$.

$F=m\left(g+\frac{v^2}{L}\right)$

Conclusion:

Substitute $15.0\text{kg}$ for $m$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1.24\text{m}/\text{s}$ for $v$ , and $14.0\text{m}$ for $L$ in the above equation to find $F$.

$\begin{array}{c}F=\left(15.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}+\frac{{\left(1.24\text{m}/\text{s}\right)}^2}{14.0\text{m}}\right)\\ =\left(15.0\text{kg}\right)\left(9.91\text{m}/{\text{s}}^{\text{2}}\right)\\ =149\text{N}\end{array}$

Therefore, the maximum tension is $149\text{N}$.

(d)

To determine

The fundamental frequency of standing wave at the time of maximum tension.

Answer to Problem 98P

The fundamental frequency is $16.3\text{Hz}$.

Explanation of Solution

Mass of object is $15.0\text{kg}$, Mass of wire is $10.0\text{g}$ length of thin wire is $14.0\text{m}$, and the maximum oscillation angle of pendulum is $6.10°$

Write the equation for the fundamental frequency of wave.

$f_1=\frac{v_1}{2L}$

Here, $f_1$ is the fundamental frequency and $v_1$ is the speed of wave at fundamental frequency.

Write the equation for $v$.

$v=\sqrt{\frac{FL}{m_w}}$

Here, $m_w$ is the mass of wire.

Rewrite the equation for $f_1$.

$\begin{array}{c}{f}_1=\frac{1}{2L}\sqrt{\frac{FL}{m_w}}\\ =\sqrt{\frac{F}{4l{m}_w}}\end{array}$

Conclusion:

Substitute $14.0\text{m}$ for $L$, $149\text{N}$ for $F$, and $10.0\text{g}$ for $m_w$ in the above equation to find $F$.

$\begin{array}{c}{f}_1=\sqrt{\frac{149\text{N}}{4\left(14.0\text{m}\right)\left(10.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)}}\\ =16.3\text{Hz}\end{array}$

Therefore, the fundamental frequency is $16.3\text{Hz}$.