A uniform taut string with two point charges $q$ and $-q$ attached to its ends passes over two massl...

To understand how the fundamental frequency $ f $ relates to the distance $ L $ between the pulleys, consider the tension in the string. This tension arises from the electrostatic force between the two charges, $ q $ and $-q$, at the ends of the string. The electrostatic force $ F_e $ is given by:$ F_e = \frac{kq_1q_2}{r^2} $where $ k = \frac{1}{4\pi\epsilon_0} $, and $ r $ is the separation between the charges, which in this setup is $ L/2 $. Substituting this, the tension $ T $ in the string becomes:$ T = \frac{4kq^2}{L^2} $The fundamental frequency $ f $ of a string is determined using the formula:$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $where $ \mu $ is the linear mass density of the string. Substitute the expression for tension $ T $ into the equation:$ f = \frac{1}{2L} \sqrt{\frac{4kq^2}{L^2\mu}} $Upon simplification, this becomes:$ f = \frac{1}{2L^2} \sqrt{\frac{4kq^2}{\mu}} $Since $ \sqrt{\frac{4kq^2}{\mu}} $ is a constant, we infer that:$ f \propto \frac{1}{L^2} $This shows that the fundamental frequency $ f $ is inversely proportional to the square of the string length $ L $.

Consider two pipes $A$ and $B$ of identical length. $A$ has one end closed and one end open. $B$ has...

1. Frequencies in Pipes:Tube A (one end closed, one end open):The frequency of the $n_A$-th harmonic in a pipe closed at one end and open at the other is given by:$f_A = n_A \frac{v_A}{4L}$where $v_A$ is the speed of sound in the gas in chamber A, $L$ is the length of the tube, and $n_A$ is an odd integer (1, 3, 5, … for the 1st, 3rd, 5th, … harmonics respectively).Tube B (both ends open):The frequency of the $n_B$-th harmonic in a pipe open at both ends is given by:$f_B = n_B \frac{v_B}{2L}$where $v_B$ is the speed of sound in the gas in chamber B, $L$ is the length of the tube, and $n_B$ is any positive integer (1, 2, 3, … for the 1st, 2nd, 3rd, … harmonics respectively).2. Speed of Sound in an Ideal Gas:The speed of sound ($v$) in an ideal gas is dependent on the temperature ($T$) and is given by:$v = \sqrt{\frac{\gamma R T}{M}}$where $\gamma$ is the adiabatic index, $R$ is the ideal gas constant, and $M$ is the molar mass of the gas.Since both chambers contain the same ideal gas (implied by "ideal gas" without specifying different types), $\gamma$, $R$, and $M$ are constants for both chambers.So, for chamber A: $v_A = \sqrt{\frac{\gamma R T_A}{M}}$And for chamber B: $v_B = \sqrt{\frac{\gamma R T_B}{M}}$3. Equating Frequencies:The problem states that the sound frequencies are the same, so $f_A = f_B$.$n_A \frac{v_A}{4L} = n_B \frac{v_B}{2L}$Now, we can cancel out the common terms. The length $L$ is identical for both tubes:$n_A \frac{v_A}{4} = n_B \frac{v_B}{2}$Multiply both sides by 4 to simplify:$n_A v_A = 2 n_B v_B$4. Substitute Speed of Sound and Solve for $T_A$:Substitute the expressions for $v_A$ and $v_B$:$n_A \sqrt{\frac{\gamma R T_A}{M}} = 2 n_B \sqrt{\frac{\gamma R T_B}{M}}$To eliminate the square roots, square both sides of the equation:$\left(n_A \sqrt{\frac{\gamma R T_A}{M}}\right)^2 = \left(2 n_B \sqrt{\frac{\gamma R T_B}{M}}\right)^2$$n_A^2 \left(\frac{\gamma R T_A}{M}\right) = (2 n_B)^2 \left(\frac{\gamma R T_B}{M}\right)$$n_A^2 \frac{\gamma R T_A}{M} = 4 n_B^2 \frac{\gamma R T_B}{M}$Now, cancel out the common factor $\frac{\gamma R}{M}$ from both sides:$n_A^2 T_A = 4 n_B^2 T_B$Finally, solve for $T_A$:$T_A = \frac{4 n_B^2}{n_A^2} T_B$

Consider two waves, which are given by $y_1(x, t)=A \sin (k x-\omega t)$ and $y_2(x, t)=\sqrt{3} A \...

Given:$ y_1(x,t) = A \sin(kx - \omega t) $$ y_2(x,t) = \sqrt{3} A \cos(kx - \omega t) $We can combine the two waves:$ y = y_1 + y_2 = A \sin(kx - \omega t) + \sqrt{3}A \cos(kx - \omega t) $Step 1: Express everything in terms of sineWe know $\cos \theta = \sin(\theta + \pi/2)$, so we can write:$ y = A \sin(kx - \omega t) + \sqrt{3}A \sin\big((kx - \omega t) + \pi/2\big) $This is a sum of two sine waves with same frequency and wavenumber but a phase difference of $\pi/2$.Step 2: Use phasor representationRepresent each wave as a phasor (vector) in the complex plane.$y_1$ corresponds to a vector of magnitude $A$ along angle $0$.$y_2$ corresponds to a vector of magnitude $\sqrt{3} A$ at angle $+\pi/2$.Step 3: Add the two phasorsLet the resultant vector be $A_s$ at angle $\phi_s$ with respect to $y_1$.The resultant is:$ \mathbf{A_s} = A + j \sqrt{3}A $where $j = i = \sqrt{-1}$.Thus:$ A_s = |A_s| = \sqrt{A^2 + (\sqrt{3}A)^2} = \sqrt{A^2(1 + 3)} = 2A $and$ \tan \phi_s = \frac{\text{Imag part}}{\text{Real part}} = \frac{\sqrt{3}A}{A} = \sqrt{3} $$ \Rightarrow \phi_s = \frac{\pi}{3} $✅Result:$ A_s = 2A , \quad \phi_s = \frac{\pi}{3} $✅Correct option:Option A$ A_s = 2A, \quad \phi_s = \frac{\pi}{3} $

Waves - IAT-IISER Physics PYQs

Q: The frequency of the whistle of a train moving with a constant speed is observed by a stationary det...

For the stationary detector:The frequency observed by the stationary detector, denoted as $ v_1 $, when the source is moving towards the detector, is given by:$ v_1 = v \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} $Where:$ v $ is the actual frequency of the train's whistle.$ v_{\text{sound}} $ is the speed of sound.$ v_{\text{source}} $ is the speed of the train (the source).For the moving detector:The frequency observed by a detector in another train moving towards the first train, denoted as $ v_2 $, is given by:$ v_2 = v \frac{v_{\text{sound}} + v_{\text{detector}}}{v_{\text{sound}} - v_{\text{source}}} $Here, $ v_{\text{detector}} $ is the speed of the second train, which is the same as $ v_{\text{source}} $ (i.e., $ v $). We need to find the ratio $\frac{v}{v_{\text{sound}}}$. To do this, we can set up a ratio of $ v_2 $ to $ v_1 $:$ \frac{v_2}{v_1} = \frac{v\left(\frac{v_{\text{sound}} + v}{v_{\text{sound}} - v}\right)}{v\left(\frac{v_{\text{sound}}}{v_{\text{sound}} - v}\right)} = \frac{v_{\text{sound}} + v}{v_{\text{sound}}} $So,$ \begin{aligned} & \frac{v_2}{v_1} = 1 + \frac{v}{v_{\text{sound}}} \\ & \frac{v}{v_{\text{sound}}} = \frac{v_2}{v_1} - 1 \end{aligned} $Therefore, the ratio $\frac{v}{v_{\text{sound}}}$ is:$ \frac{v}{v_{\text{sound}}} = \frac{v_2 - v_1}{v_1} $

01

PYQ 2022

medium

physics ID: iat-iise

A uniform taut string with two point charges and attached to its ends passes over two massless pullies kept distance apart as shown in the fgure. If is the fundamental frequency of the part of the string over pullies, which of the following statements is correct?

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02

PYQ 2023

easy

physics ID: iat-iise

The frequency of the whistle of a train moving with a constant speed is observed by a stationary detector on the platform to be . The frequency of the same whistle is detected to be inside another train moving on a parallel track, at a speed towards the frst train. If the speed of sound in air is taken to be ound, what is the ratio ound?

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03

PYQ 2025

easy

physics ID: iat-iise

Consider two pipes and of identical length. has one end closed and one end open. has both ends open. Each tube is immersed in a closed chamber of ideal gas having volume . The chamber containing tube is at temperature and the chamber containing tube is at temperature . The sound frequencies corresponding to the -th harmonic in tube and the -th harmonic in tube are the same. What is the relation between the temperatures and ?

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$ $

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$ $

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$ $

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$ $

04

PYQ 2025

medium

physics ID: iat-iise

Consider two waves, which are given by and, where is the wave number and is the angular frequency. The amplitude of the resultant waveform obtained by the superposition of the two waves is and its phase difference with is . What are and ?

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$ $

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$ $

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$ $

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$ $

Waves

High-Yield Trend

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4 MCQs

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Waves

High-Yield Trend

Questions 4 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Questions
4 MCQs