91Ó°ÊÓ

(a) When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note B flat, which has a frequency of 466 Hz, how much time does it take the person’s vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords?

(b) When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that young humans can hear has a period of 50.0 micro second. What are the frequency and angular frequency of the vibrating eardrum for this sound?

(c) When light having vibrations with angular frequency ranging from$\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$to$\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{15}}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$ strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light?

(d) High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumours, a frequency of around 5.0 MHz is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

Step by step solution

01

Calculate the angular frequency of the chordsa)

• The given frequency is f = 466 Hz

Also, the relation between time period and frequency is:

$\mathit{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}$

Therefore, the time period can be calculated as:

$\begin{array}{l}\mathbf{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{f}}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{466}}\\ \mathbf{=}\mathbf{2}\mathbf{.}\mathbf{15}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{s}\end{array}$

The angular frequency is related to the time period as:

$\mathit{Ó\neg }\mathbf{=}\mathbf{2}\mathit{Ï€}\mathit{f}$

Therefore, angular frequency

$\begin{array}{l}\mathrm{Ó\neg }=2\mathrm{Ï€}\mathrm{f}\\ =2\mathrm{Ï€}×\left(466\right)\\ =2.93×{10}^{3}rad/\mathrm{s}\end{array}$

It takesrole="math" localid="1668077051761" $2.15x{10}^{-3}s$ time for the person’s vocal cords to vibrate through one complete cycle, and the angular frequency of the cord is $2.93×{10}^{3}rad/s$.

02

Calculate the frequency and angular frequency of the eardrumb)

• We have time$\mathit{T}\mathbf{=}\mathbf{50}\mathit{x}{{\mathbf{10}}^{\mathbf{-}}}^{\mathbf{6}}\mathbf{}\mathit{s}$

We can use,

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{T}}$

to find the frequency.

Therefore, the frequency can be calculated as:

$\begin{array}{l}f=\frac{1}{T}\\ =\frac{1}{50×{10}^{-6}}\\ =2×{10}^{4}Hz\end{array}$

Now, use frequency to find angular frequency as:

$\begin{array}{r}\mathrm{Ó\neg }=2\mathrm{Ï€}f\\ =2\mathrm{Ï€}×\left(2×{10}^4\mathrm{Hz}\right)\\ =1.26×{10}^5\mathrm{rad}/\mathrm{s}\end{array}$

The frequency is $2×{10}^4\mathrm{Hz}$and angular frequency is $1.26×{10}^{5}rad/s$of the vibrating eardrum for this sound.

03

Calculate limits of period and frequency of lightc)

We have,

• ${\mathit{Ó\neg }}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.7}\mathbf{×}{\mathbf{10}}^{\mathbf{15}}\mathbf{rad}\mathbf{/}\mathbf{s}$,
• ${\mathit{Ó\neg }}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.7}\mathbf{×}{\mathbf{10}}^{\mathbf{15}\mathbf{}}\mathbf{rad}\mathbf{/}\mathbf{s}$
  

Also,

$\mathit{f}\mathbf{=}\frac{\mathbf{Ó\neg }}{\mathbf{2}\mathbf{Ï€}}$

Therefore, to determine the limits of the frequency we can use the formula as:

$\begin{array}{r}{f}_1=\frac{{\mathrm{Ó\neg }}_1}{2\mathrm{Ï€}}\\ =\frac{2.7×{10}^{15}\mathrm{rad}/\mathrm{s}}{2\mathrm{Ï€}}\\ =4.3×{10}^{14}\mathrm{Hz}\\ f_2=\frac{{\mathrm{Ó\neg }}_2}{2\mathrm{Ï€}}\\ =\frac{4.7×{10}^{15}\mathrm{rad}/\mathrm{s}}{2\mathrm{Ï€}}\\ =7.5×{10}^{14}\mathrm{Hz}\end{array}$

Now, to determine the limits of the period, use the above values of frequencies,

$\begin{array}{r}{T}_1=\frac{1}{f_1}\\ =\frac{1}{4.3×{10}^{14}}\\ =2.3×{10}^{−15}\mathrm{s}\\ T_2=\frac{1}{f_2}\\ =\frac{1}{7.5×{10}^{14}}\\ =1.3×{10}^{−15}\mathrm{s}\end{array}$

Limits of the period are $1.3x{10}^{-15}s$to $2.3x{10}^{-15}s$and frequency is $4.3×{10}^{14}Hz$to$7.5×{10}^{14}Hz$ of this light.

04

Calculate the period and angular frequency of the molecular vibrations caused by the pulse of soundd)

We have frequency

• $\mathit{f}\mathbf{=}\mathbf{5}\mathbf{×}{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathit{H}\mathit{z}$
  

The time period can be calculated as:

$\begin{array}{l}T=\frac{1}{f}\\ =\frac{1}{5×{10}^6}\\ =2×{10}^{-7}s\end{array}$

We can find angular frequencyusing the above value of frequency as:

$\begin{array}{r}\mathrm{Ó\neg }=2\mathrm{Ï€}f\\ =2\mathrm{Ï€}\left(5×{10}^6\mathrm{Hz}\right)\\ =3.1×{10}^7\mathrm{rad}/\mathrm{s}\end{array}$

The period is $2×{10}^{-7}s$and angular frequency of the molecular vibrations caused by this pulse of sound is $3.1x{10}^{7}rad/s$.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!