Съдържание

INPhO 2019, Теория — Задача 1. Propagation of sound

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2019

Indian National Physics Olympiad – 2019 · 3 февруари 2019 г. · 9 т.

Внимание: Бележка към темата

The complete retained question booklet includes summary answer boxes, eighteen blank detailed-answer pages and three blank graph sheets. All matching archived HBCSE key content is retained, including both full handwritten graph sheets and accepted alternatives/ranges. Printed exponent, sign, factor, notation and unit inconsistencies remain literal with labelled source notes.

Native cover and instructions

Indian National Physics Olympiad – 2019

Date: 03 February 2019. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,9\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ . Extra sheets attached: [blank box]. Centre (e.g. Kota): [blank line].

(Do not write below this line)

This booklet consists of 29 pages (excluding this page) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	9
2	11
3	12
4	7
5	9
6	14
7	13
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover

Solutions of Indian National Physics Olympiad – 2019

The native key cover repeats the nine instructions, twelve constant rows, seven-question score table totalling 75 marks, roll-number form and HBCSE address from the question cover. It prints “20 pages (excluding this page)”; the retained key contains twenty-one PDF pages, including two appended handwritten graph sheets.

Условие

In the lower part of the earth’s atmosphere, the temperature decreases with increase of height. Choose the origin of the coordinate system at the ground level with the $y$ -axis vertically upward and the $x$ -axis horizontal. We assume a linear decrease of temperature such that the temperature at a height $y$ from the ground level is

T(y)=T_0(1-by)

where $T_0$ is the temperature at the ground level. The constant $b=0.023\ \mathrm{km^{-1}}$ . We consider the propagation of sound in the $x$ - $y$ plane. Ignore any attenuation, reflection, and diffraction of sound.

Native answer booklet: PDF pages10–26 and30 (printed pages9–25 and29) are eighteen blank Detailed Answers pages with a “Que. No.” box and, on alternate printed pages, a “Last four digits of Roll No.:” field. The complete original remains available; its first blank answer-page template is retained here.

Whole first blank native Detailed Answers sheet, printed page9, blank question-number box and complete answer frame.

a) If $v_0$ is the speed of sound at the ground level, obtain an expression for the speed of sound $v(y)$ at height $y$ , in terms of $v_0$ and $b$ .

Summary answer box: $v(y)=$ [blank]. [1 т.]

b) Suppose sound propagates from the origin with an initial angle $\theta_0$ with the $x$ -axis. Obtain an expression for the angle $\theta$ made by the direction of propagation of sound with the horizontal at height $y$ , in terms of $\theta_0$ and $b$ .

Summary answer box: $\theta=$ [blank]. [2 т.]

c) Obtain an expression for the $x$ and $y$ coordinates of a point on the path of propagation as functions of $\theta$ .

Summary answer boxes: $x=$ [blank]; $y=$ [blank]. [3 т.]

d) For this part assume the direction of propagation of sound to be horizontal at the origin. For the case where $y$ is of the order $100\ \mathrm m$ or less, obtain an approximated expression relating $x$ and $y$ i.e. $y(x)$ . Obtain $x$ for $y=2.00\ \mathrm m$ .

Summary answer boxes: $y(x)=$ [blank]; Value of $x=$ [blank].

Detailed answers can be found on page numbers: [blank]. [3 т.]

Решение

(a)

Speed of sound is

v=\sqrt{\frac{\gamma RT}{m}}\qquad (1.1)

where $m$ and $\gamma$ are the molar mass and adiabatic index of the gas respectively.

v(y)=\sqrt{\frac{\gamma RT(y)}m}=\sqrt{\frac{\gamma RT_0}m}\sqrt{1-by}=v_0\sqrt{1-by}\qquad (1.2)

Here $v_0=\sqrt{\gamma RT_0/m}$ is the speed of the sound at ground level.

(b)

Consider the propogation of sound ‘ray’ from one medium to the other.

Entire native sound-ray refraction sketch with interface, dashed normal, incident/transmitted arrows, i1/i2 and theta1/theta2.

From Snell’s law

\frac{\sin i_2}{\sin i_1}=\frac{v_2}{v_1}\qquad (1.3)

\frac{\sin(\pi/2-\theta_2)}{\sin(\pi/2-\theta_1)}=\frac{\cos\theta_2}{\cos\theta_1}=\frac{v_2}{v_1}\qquad (1.4)

\frac{\cos\theta}{\cos\theta_0}=\frac{v(y)}{v_0}=\sqrt{1-by}\qquad (1.5)

\theta=\cos^{-1}\left[\cos\theta_0\sqrt{1-by}\right]\qquad (1.6)

(c)

y=\frac1b\left(1-\frac{\cos^2\theta}{\cos^2\theta_0}\right)\qquad (1.7)

\frac{dy}{dx}=\tan\theta\qquad (1.8)

dx=\frac{2\cos\theta\sin\theta}{\tan\theta\,b\cos^2\theta_0}\,d\theta\qquad (1.9)

x=\int_0^x dx=\frac1{2b\cos^2\theta_0}\left[2(\theta-\theta_0)+\sin2\theta-\sin2\theta_0\right]\qquad (1.10)

(d)

For $y\le100\ \mathrm m$ and $\theta_0=0$ , Eq. (1.6) gives $\theta\approx3^\circ$ (very small).

y\approx\frac1b\left[1-\left(1-\frac{\theta^2}{2}\right)^2\right]=\frac1b\theta^2\qquad (1.11)

x\approx\frac1{2b}(2\theta+2\theta)=\frac{2\theta}b\qquad (1.12)

x^2=\frac{4y}b\qquad (1.13)

x(y=2)\approx590\ \mathrm m\qquad (1.14)

Accepted range: $585$ – $600\ \mathrm m$ .

Оригинал в Архива: INPhO2019-Question.pdf · решение от архива: INPhO2019-Solution-20190219.pdf