Chapter 12, Problem 81P

(a)

To determine

The wavelength of the fundamental mode of vibration.

Answer to Problem 81P

The wavelength of the fundamental mode is $1.28\text{m}$.

Explanation of Solution

The length of the string is $64.0\text{cm}$, the tension on the string is $133\text{N}$ and the fundamental frequency is $110.0\text{Hz}$.

Write the expression for wavelength of the fundamental mode

$\lambda =2L$                                                            (I)

Here, $\lambda$ is the wavelength of the fundamental mode and $L$ is the length of the string.

Substitute $64.0\text{cm}$ for $L$ in (I) to find $\lambda$

$\begin{array}{c}\lambda =2\left(64.0\text{cm}\right)\\ =2\left(64.0\times {10}^{-2}\text{m}\right)\\ =1.28\text{m}\end{array}$

Thus, the wavelength of the fundamental mode is $1.28\text{m}$.

(b)

To determine

The speed of the wave.

Answer to Problem 81P

The speed of the wave is $141{\text{ms}}^{-1}$.

Explanation of Solution

Write the expression for speed of the wave

$v=\lambda f$                                                            (II)

Here, $v$ is the wave speed and $f$ is the frequency.

Substitute $1.28\text{m}$ for $\lambda$ and $110.0\text{Hz}$ for $f$ in (II) to find $v$

$\begin{array}{c}v=\left(1.28\text{m}\right)\left(110.0\text{Hz}\cdot \frac{1{\text{s}}^{-1}}{1\text{Hz}}\right)\\ =141{\text{ms}}^{-1}\end{array}$

Thus, the speed of the wave is $141{\text{ms}}^{-1}$.

(c)

To determine

The linear mass density of the string.

Answer to Problem 81P

The linear mass density of the string is $6.71{\text{ gm}}^{-1}$.

Explanation of Solution

Write the expression for velocity of wave propagation on the string

$v=\sqrt{\frac{T}{\mu }}$                                                                   (III)

Here, $T$ is the tension on the wire and $\mu$ is the linear mass density.

Rearrange for $\mu$

$\mu =\frac{T}{v^2}$                                                                    (IV)

Substitute $141{\text{ms}}^{-1}$ for $v$ and $133\text{N}$ for $T$ in (IV) to find $\mu$

$\begin{array}{c}\mu =\frac{133\text{N}}{{\left(141{\text{ms}}^{-1}\right)}^2}\\ =\frac{\left(133{\text{kgs}}^{-2}\right)}{{\left(141{\text{ms}}^{-1}\right)}^2}\\ =6.71\times {10}^{-3}\text{}{\text{kgm}}^{-1}\\ =6.71{\text{ gm}}^{-1}\end{array}$

Thus, the linear mass density of the string is $6.71{\text{ gm}}^{-1}$.

(d)

To determine

The maximum speed of a point on the vibrating string.

Answer to Problem 81P

The maximum speed of a point on the vibrating string $1.59{\text{ms}}^{-1}$.

Explanation of Solution

The maximum amplitude on the string is $2.30\text{mm}$ and the fundamental frequency is $110.0\text{Hz}$.

Write the expression for maximum speed

$v_{\text{m}}=\omega A$                                                             (V)

Here, $v_{\text{m}}$ is the maximum speed of a point, $\omega$ is the angular frequency and $A$ is the maximum amplitude.

Substitute for $\omega$

$v_{\text{m}}=2\pi fA$                                                                  (VI)

Here, $f$ is the fundamental frequency.

Substitute $110.0\text{Hz}$ for $f$ and $2.30\text{mm}$ for $A$ in (VI) to find $v_{\text{m}}$

$\begin{array}{c}{v}_{\text{m}}=2\pi \left(110.0\text{Hz}\right)\left(2.30\text{mm}\right)\\ =2\pi \left(110.0{\text{s}}^{-1}\right)\left(2.30\times {10}^{-3}\text{m}\right)\\ =1588\times {10}^{-3}{\text{ ms}}^{-1}\\ \approx 1.59{\text{ms}}^{-1}\end{array}$

Thus, the linear mass density of the string is $1.59{\text{ms}}^{-1}$.

(e)

To determine

The frequency of the sound wave.

Answer to Problem 81P

The frequency of the sound wave in air is $110.0\text{Hz}$.

Explanation of Solution

The frequency of the vibrating string is $110.0\text{Hz}$.

The vibrating string puts the air molecules around it into vibratory motion. The frequency of vibration of the air molecules is same as the frequency of the vibrating string.

Hence the frequency of the sound wave in air is $110.0\text{Hz}$.

(f)

To determine

The wavelength of sound in air medium at $20.0°\text{C}$ .

Answer to Problem 81P

The wavelength of sound at $20.0°\text{ C}$ is $3.120\text{m}$.

Explanation of Solution

The air temperature in kelvin is $\left(20.0+273.15\right)\text{K}$.

Rearrange (II)

$\lambda =\frac{v}{f}$                                                                  (VII)

Here, $v$ is the velocity of sound in air, $\lambda$ is the wavelength in air and $f$ is the frequency in air.

Write the expression for temperature dependence of velocity of sound

$v=v_0\sqrt{\frac{\theta }{{\theta }_0}}$                                                                (VIII)

Here, $v_0$ is the velocity of sound at absolute temperature $\theta$ is the given temperature in kelvin and ${\theta }_0$ is the absolute temperature in kelvin

Substitute $343{\text{ms}}^{-1}$ for $v_0$, $\left(20.0+273.15\right)\text{K}$ for $\theta$ , and $273.15\text{K}$ for ${\theta }_0$ in (VIII) to find $v$.

$\begin{array}{c}v=343{\text{ms}}^{-1}\sqrt{\frac{\left(20.0+273.15\right)\text{K}}{273.15\text{K}}}\\ =343.2{\text{ms}}^{-1}\end{array}$

Substitute $343.2{\text{ms}}^{-1}$ for $v$ and $110.0\text{Hz}$ for $f$ in (VII) to find $\lambda$

$\begin{array}{c}\lambda =\frac{343.2{\text{ms}}^{-1}}{110.0\text{Hz}}\\ =\frac{343.2{\text{ms}}^{-1}}{110.0\text{Hz}}\\ =3.120\text{m}\end{array}$

Thus, the wavelength of sound at $20.0°\text{ C}$ is $3.120\text{m}$.