Category: HC Verma Part 1: Waves & Optics
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HC Verma Chapter 16 Problem 26 — end correction in closed pipe
Problem Statement A closed pipe resonates at $L_1=16.5$ cm (1st) and $L_2=49.5$ cm (2nd) for a 512 Hz fork. Find end correction. Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on fundamental physical principles. The key…
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HC Verma Chapter 16 Problem 24 — speed of sound in steel
Problem Statement Solve the kinematics problem: Speed of sound in steel. $Y=2\times10^{11}$ Pa, $\rho=7800$ kg/m$^3$. $v=\sqrt{Y/\rho}$ for solids Step 1: $v=\sqrt{Y/\rho}=\sqrt{2\times10^{11}/7800}=\sqrt{2.564\times10^7}\approx5064$ m/s. $$\boxed{v\approx5064\text{ m/s}}$$ Given Information Initial velocity $u$ (or $v_0$) Acceleration $a$ (constant unless stated otherwise) Time $t$ or distance $s$ as given Physical Concepts & Formulas Kinematics describes motion without reference to its…
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HC Verma Chapter 16 Problem 22 — intensity level at distance from point source
Problem Statement A small speaker produces 40 W. Find intensity level at 2 m. $I_0=10^{-12}$ W/m$^2$. Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on fundamental physical principles. The key is to identify which conservation law…
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HC Verma Chapter 16 Problem 23 — pressure amplitude of sound wave
Problem Statement Solve the oscillation/wave problem: A sound wave of frequency 500 Hz, displacement amplitude $10^{-5}$ m in air. $\rho=1.29$ kg/m$^3$, $v=330$ m/s. Find pressure amplitude. $p_0=\rho v\omega s_0$ Step 1: $p_0=\rho v\omega s_0=1.29\times330\times(2\pi\times500)\times10^{-5}$. Step 2: $=1.29\times330\times3142\times10^{-5}=1.29\times330 Given Information Mass $m$ and spring constant $k$ (or equivalent), or wave parameters Initial conditions (amplitude $A$, phase…
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HC Verma Chapter 16 Problem 20 — minimum distance to hear echo
Problem Statement What is the minimum distance from a wall for an echo to be heard distinctly? Persistence of hearing = 0.1 s. $v=340$ m/s. Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on fundamental physical…
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HC Verma Chapter 16 Problem 21 — sound intensity at distance
Problem Statement A point source emits 40 W of acoustic power. Find intensity at 10 m distance. Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on fundamental physical principles. The key is to identify which conservation…
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HC Verma Chapter 16 Problem 19 — echo time calculation
Problem Statement A man stands 330 m from a wall and claps. After how long does he hear the echo? $v=330$ m/s. Given Information All quantities, constants, and constraints stated in the problem above Physical constants used as needed (see Concepts section) Physical Concepts & Formulas This problem draws on fundamental physical principles. The key…
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HC Verma Chapter 16 Problem 17 — Doppler effect moving observer toward source
Problem Statement Solve the oscillation/wave problem: Observer moves toward stationary 500 Hz source at 30 m/s. $v=330$ m/s. Observer approaching: $v+v_o$ in numerator Step 1: $f’=f\frac{v+v_o}{v}=500\times\frac{360}{330}=545.5$ Hz. $$\boxed{f’\approx545\text{ Hz}}$$ Given Information Mass $m$ and spring constant $k$ (or equivalent), or wave parameters Initial conditions (amplitude $A$, phase $\phi$) as given Physical Concepts & Formulas Simple…
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HC Verma Chapter 16 Problem 18 — Doppler — observer moving away from source
Problem Statement Solve the oscillation/wave problem: Observer moves away from 500 Hz source at 30 m/s. $v=330$ m/s. Observer receding: $v-v_o$ in numerator Step 1: $f’=f\frac{v-v_o}{v}=500\times\frac{300}{330}=454.5$ Hz. $$\boxed{f’\approx455\text{ Hz}}$$ Given Information Mass $m$ and spring constant $k$ (or equivalent), or wave parameters Initial conditions (amplitude $A$, phase $\phi$) as given Physical Concepts & Formulas Simple…
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HC Verma Chapter 16 Problem 16 — Doppler effect moving source away
Problem Statement Solve the oscillation/wave problem: Same source (500 Hz) moves away from observer at 30 m/s. $v=330$ m/s. Source receding: $v+v_s$ in denominator Step 1: $f’=f\frac{v}{v+v_s}=500\times\frac{330}{360}=458.3$ Hz. $$\boxed{f’\approx458\text{ Hz}}$$ Given Information Mass $m$ and spring constant $k$ (or equivalent), or wave parameters Initial conditions (amplitude $A$, phase $\phi$) as given Physical Concepts & Formulas…