Q.Equation of a plane progressive wave is given by $y = 0.6\sin 2\pi\left(t - \dfrac{x}{2}\right)$. On reflection from a denser medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is

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1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Two identical strings A and B of same material are having tensions $T_A$ and $T_B$ respectively. If their fundamental frequencies are 450 Hz and 300 Hz, then $T_A/T_B$ is (A) $\dfrac{4}{9}$ (B) $\dfrac{2}{9}$ (C) $\dfrac{6}{4}$ (D) $\dfrac{9}{4}$

   ›Reveal solutionSolution

   Tests how fundamental frequency of a stretched string depends on tension. Answer: $T_A/T_B=9/4$.

   Concept and Intuition

   The fundamental frequency of a vibrating string is $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, where $L$ is length and $\mu$ is mass per unit length. Since the strings are identical (same material, same $L$ and $\mu$), any difference in frequency between them must come entirely from the difference in tension — and because frequency depends on the square root of tension, a modest frequency ratio corresponds to a larger tension ratio.

   Step-by-Step Solution

   1. For identical strings, $f\propto\sqrt{T}$ (with $L,\mu$ the same for both). …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A wire of density $9\times10^3\ kg\,m^{-3}$ is stretched between two clamps of 1 m apart and is subjected to an extension of $4.9\times10^{-4}\ m$. The lowest frequency of transverse vibrations in the wire is (Young's modulus of the material of the wire $Y = 9\times10^{10}\ Nm^{-2}$) (A) 40 Hz (B) 35 Hz (C) 30 Hz (D) 25 Hz

   ›Reveal solutionSolution

   Combines Young's modulus (to get the tension-per-unit-area, i.e. stress) with the standing-wave formula for a stretched string to find the fundamental frequency — cleverly, the cross-sectional area cancels out entirely. Answer: 35 Hz.

   Concept and Intuition

   The lowest ("fundamental") frequency of transverse vibration of a wire clamped at both ends is

   $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

   where $T$ is the tension and $\mu = \rho A$ is mass per unit length. Normally we'd need to know $T$ and $A$ separately. But here we're given the extension produced, which lets us find the stress (force per unit area) via Young's modulus — and since $T = (\text{stress})\times A$ while $\mu = \rho A$, the ratio $T/\mu = \text{stress}/\rho$ is independent of the (unknown) area $A$! This is the elegant trick: we never need to know the wire's actual radius.

   Step-by-Step Solution

   1. Strain: $\dfrac{\Delta L}{L} = \dfrac{4.9\times10^{-4}}{1} = 4.9\times10^{-4}$.
   2. Stress from Young's modulus: $\text{stress} = Y\times\text{strain} = 9\times10^{10}\times4.9\times10^{-4} = 4.41\times10^{7}\ N/m^2$.
   3. Since $T = \text{stress}\times A$ and $\mu=\rho A$: …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The mass and length of a string are 10 g and 100 cm respectively. If the tension in the string is increased from 400 N to 900 N, then the increase in the frequency of transverse vibration of the string is (A) 200 Hz (B) 150 Hz (C) 50 Hz (D) 100 Hz

   ›Reveal solutionSolution

   Tests the formula for the natural (fundamental) frequency of a stretched string, $f=\frac{1}{2L}\sqrt{T/\mu}$, and how it changes when tension changes.

   Concept and Intuition

   A stretched string's vibration frequency depends on how taut it is (tension $T$, which stiffens it and raises pitch) and how heavy it is per unit length ($\mu$, which slows the wave and lowers pitch), scaled by its length. Frequency grows only as $\sqrt T$, not linearly, so equal jumps in tension don't give equal jumps in frequency — that's exactly what's being tested here.

   Step-by-Step Solution

   1. Linear mass density: $\mu = \dfrac{m}{L} = \dfrac{10\times10^{-3}\,\text{kg}}{1\,\text{m}} = 0.01\,\text{kg/m}$.
   2. Fundamental frequency: $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, with $L=1\,m$ fixed throughout.
   3. At $T_1=400\,N$: $f_1 = \dfrac12\sqrt{\dfrac{400}{0.01}} = \dfrac12\sqrt{40000} = \dfrac12(200) = 100\,\text{Hz}$. …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The speed of a stationary wave represented by the equation $y=0.7\sin\left(\dfrac{7\pi}{4}x\right)\cos(350\pi t)$ is (In the given equation x and y are in metre and t is in second) (A) $100\ \mathrm{m\,s^{-1}}$ (B) $150\ \mathrm{m\,s^{-1}}$ (C) $160\ \mathrm{m\,s^{-1}}$ (D) $200\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   A standing wave is a superposition of two travelling waves of the same speed; reading off $k$ and $\omega$ from the given equation and computing $v=\omega/k$ gives $200\ \mathrm{m\,s^{-1}}$.

   Concept and Intuition

   Although a stationary (standing) wave itself doesn't propagate energy along the string the way a travelling wave does, it is built from two identical travelling waves moving in opposite directions, each with speed $v=\omega/k$ (angular frequency over wave number). So we can still extract "the speed of the wave" by comparing the given equation to the standard form $y = A\sin(kx)\cos(\omega t)$.

   Step-by-Step Solution

   1. Standard form: $y = A\sin(kx)\cos(\omega t)$, where $k$ is the wave number and $\omega$ the angular frequency.
   2. Comparing with $y = 0.7\sin\left(\dfrac{7\pi}{4}x\right)\cos(350\pi t)$: $k = \dfrac{7\pi}{4}\ \mathrm{rad\,m^{-1}}$, $\omega = 350\pi\ \mathrm{rad\,s^{-1}}$. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.When a stretched wire of fundamental frequency f is divided into three segments, the fundamental frequencies of these three segments are $f_1$, $f_2$ and $f_3$ respectively. Then the relation among $f_1$, $f_2$, $f_3$ and f is (Assume tension is constant) (A) $\sqrt{f} = \sqrt{f_1} + \sqrt{f_2} + \sqrt{f_3}$ (B) $f = f_1 + f_2 + f_3$ (C) $\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dfrac{1}{f_3}$ (D) $\dfrac{1}{\sqrt{f}} = \dfrac{1}{\sqrt{f_1}} + \dfrac{1}{\sqrt{f_2}} + \dfrac{1}{\sqrt{f_3}}$

   ›Reveal solutionSolution

   This tests how the fundamental frequency of a stretched string scales with length when the wire is physically cut into pieces under the same tension. The lengths add, so the RECIPROCALS of the frequencies add: $1/f = 1/f_1+1/f_2+1/f_3$.

   Concept and Intuition

   The fundamental frequency of a stretched wire fixed at both ends is $f = \dfrac{v}{2L}$, where $v=\sqrt{T/\mu}$ is the transverse wave speed set by the tension $T$ and the linear mass density $\mu$. When the SAME wire (same $\mu$) is cut into three pieces and each piece is stretched with the SAME tension $T$, the wave speed $v$ in every piece is identical — only the length changes. So frequency is inversely proportional to length, and it's the LENGTHS that add up to give back the original wire, not the frequencies directly.

   Step-by-Step Solution

   1. Original wire: $f = \dfrac{v}{2L} \Rightarrow L = \dfrac{v}{2f}$.
   2. Segment $i$: $f_i = \dfrac{v}{2L_i} \Rightarrow L_i = \dfrac{v}{2f_i}$, for $i=1,2,3$ (same $v$ since tension & mass density unchanged).
   3. The three segments together make up the whole wire: $L_1+L_2+L_3 = L$.
   4. Substitute: $\dfrac{v}{2f_1}+\dfrac{v}{2f_2}+\dfrac{v}{2f_3} = \dfrac{v}{2f}$. …

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6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A steel wire of length 81 cm has a mass of $5 \times 10^{-3}$ kg. If the wire is under a tension of 50 N, then the speed of transverse waves on the wire is (A) $100\ \mathrm{m\,s^{-1}}$ (B) $105\ \mathrm{m\,s^{-1}}$ (C) $90\ \mathrm{m\,s^{-1}}$ (D) $60\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   The transverse wave speed on a stretched string depends on tension and linear mass density; here it works out to $90\ \mathrm{m/s}$.

   Concept and Intuition

   A transverse wave on a taut string travels faster when the string is under greater tension (a stronger restoring force) and slower when the string is heavier per unit length (more inertia to accelerate). This is captured in $v=\sqrt{T/\mu}$, where $\mu$ is the linear mass density.

   Step-by-Step Solution

   1. Length $L=81\ \mathrm{cm}=0.81\ \mathrm m$, mass $m=5\times10^{-3}\ \mathrm{kg}$.
   2. Linear mass density: $\mu = \dfrac{m}{L} = \dfrac{5\times10^{-3}}{0.81}\ \mathrm{kg/m}$. …

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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.The equation of a transverse wave propagating along a stretched string of length 80 cm is $y = 1.5\sin\{(5\times10^{-3}x) + 20t\}$, here 'x' and 'y' are in cm and the time 't' is in second. If the mass of the string is 3 g, then the tension in the string is (A) 12 N (B) 4 N (C) 6 N (D) 8 N

   ›Reveal solutionSolution

   Reading off $\omega$ and $k$ from the wave equation gives the wave speed; combined with the string's linear mass density, $v=\sqrt{T/\mu}$ gives the tension as $6\ \mathrm N$.

   Concept and Intuition

   A transverse wave on a string is described by $y=A\sin(kx+\omega t)$, where $k$ is the wave number and $\omega$ the angular frequency; the wave (phase) speed is $v=\omega/k$. This speed is set by the string's mechanical properties through

   $v = \sqrt{\frac{T}{\mu}}$

   where $T$ is the tension and $\mu$ the mass per unit length. So once we read $v$ off the wave equation and compute $\mu$ from the given mass and length, we can solve for $T$.

   Step-by-Step Solution

   1. Compare $y=1.5\sin\{(5\times10^{-3}x)+20t\}$ with $y=A\sin(kx+\omega t)$: $k = 5\times10^{-3}\ \mathrm{rad/cm}$ (since $x$ is in cm), $\omega=20\ \mathrm{rad/s}$.
   2. Convert $k$ to SI (rad/m): $k = 5\times10^{-3}\ \mathrm{cm^{-1}} \times 100\ \mathrm{cm/m} = 0.5\ \mathrm{m^{-1}}$.
   3. Wave speed: $v = \omega/k = 20/0.5 = 40\ \mathrm{m/s}$. …

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8. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.Two vibrating strings of same length, same cross-sectional area and stretched to same tension are made of material with densities $\rho$ and $2\rho$. Each string is fixed at both ends. If $V_1$ and $V_2$ are speeds of transverse waves in the strings with densities $\rho$ and $2\rho$ respectively, then $\dfrac{V_1}{V_2}$ is (A) $1:2$ (B) $2:1$ (C) $\sqrt{2}:1$ (D) $1:\sqrt{2}$

   ›Reveal solutionSolution

   This tests the transverse-wave-speed formula $v=\sqrt{T/\mu}$ and how linear mass density $\mu=\rho A$ scales with the material's volume density when cross-section and tension are fixed.

   Concept and Intuition

   The speed of a transverse wave on a stretched string depends on the string's tension and its linear mass density $\mu$ (mass per unit length), not directly on the material's volume density $\rho$. But since $\mu=\rho A$ (density times cross-sectional area), and here both strings share the same length, area, and tension, a denser material simply means a proportionally larger $\mu$ — and wave speed decreases as $1/\sqrt{\mu}$, i.e. as $1/\sqrt\rho$.

   Step-by-Step Solution

   1. Wave speed on a string: $v=\sqrt{\dfrac{T}{\mu}}$, with $\mu=\rho A$.
   2. String 1 has density $\rho$: $\mu_1=\rho A$, so $V_1=\sqrt{\dfrac{T}{\rho A}}$.
   3. String 2 has density $2\rho$: $\mu_2=2\rho A$, so $V_2=\sqrt{\dfrac{T}{2\rho A}}$. …

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9. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The speed of a wave on a string is $150\ \text{ms}^{-1}$ when the tension is 120 N. The percentage increase in the tension in order to raise the wave speed by 20% is (A) 44 (B) 40 (C) 22 (D) 20

   ›Reveal solutionSolution

   This tests how wave speed on a string scales with tension: $v\propto\sqrt{T}$, so a percentage change in $v$ requires roughly double that percentage change in $T$.

   Concept and Intuition

   The speed of a transverse wave on a stretched string is $v=\sqrt{T/\mu}$, where $\mu$ is the mass per unit length. Since $v$ depends on the square root of tension, a modest fractional increase in speed requires a larger fractional increase in tension.

   Step-by-Step Solution

   1. Given $v_1=150\ \text{ms}^{-1}$ at $T_1=120$ N.
   2. We want $v_2=1.20\,v_1=180\ \text{ms}^{-1}$.
   3. From $v\propto\sqrt{T}$: $\dfrac{T_2}{T_1}=\left(\dfrac{v_2}{v_1}\right)^2=(1.20)^2=1.44$. …

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10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.A string of length 'L' is stretched by $\frac{L}{20}$ and the speed of transverse waves along it is 'v'. The speed of wave when it is stretched by $\frac{L}{10}$ will be (assume that Hooke's law is applicable) (A) 2 v (B) $\frac{v}{\sqrt{2}}$ (C) $v\sqrt{2}$ (D) 4 v

    ›Reveal solutionSolution

    Since tension is proportional to the elastic extension (Hooke's law) and wave speed scales as $\sqrt{T}$, doubling the extension scales the speed by $\sqrt2$.

    Concept and Intuition

    The speed of a transverse wave on a stretched string is $v=\sqrt{T/\mu}$, where $T$ is tension and $\mu$ is mass per unit length. When a string obeys Hooke's law, the tension produced by stretching is directly proportional to the amount of extension: $T = YA\cdot\dfrac{\Delta L}{L}$ for fixed original length $L$, area $A$, and Young's modulus $Y$. Since $\Delta L$ is small, $\mu$ (mass/length) is essentially unaffected by the extra stretch, so $v\propto\sqrt{\Delta L}$.

    Step-by-Step Solution

    1. Initially, extension $\Delta L_1 = \dfrac{L}{20}$, giving speed $v$. So $v \propto \sqrt{\Delta L_1}$.
    2. New extension: $\Delta L_2 = \dfrac{L}{10} = 2\times\dfrac{L}{20} = 2\Delta L_1$. …

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11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.Small amplitude progressive wave in a stretched string has a speed of $100\ \mathrm{cms}^{-1}$, and frequency 100 Hz. The phase difference between two points 2.75 cm apart on the string, in radians, is (A) 0 (B) $11\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{3\pi}{8}$

    ›Reveal solutionSolution

    The path difference (2.75 cm) is 2.75 wavelengths (since $\lambda=1$ cm here), so the phase difference is $2\pi\times2.75=\dfrac{11\pi}{2}$ radians.

    Concept and Intuition

    For a travelling wave, phase difference between two points separated by a path difference $\Delta x$ is $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x$. This directly compares the separation to the wavelength: every full wavelength of separation corresponds to a full $2\pi$ of phase.

    Step-by-Step Solution

    1. Wave speed $v=100\ \mathrm{cm\,s^{-1}}$, frequency $f=100\ \mathrm{Hz}$.
    2. Wavelength: $\lambda=\dfrac{v}{f}=\dfrac{100}{100}=1\ \mathrm{cm}$.
    3. Path difference given: $\Delta x=2.75\ \mathrm{cm}=2.75\lambda$.
    4. Phase difference: $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x=2\pi(2.75)=5.5\pi=\dfrac{11\pi}{2}\ \mathrm{rad}$.

    Common Mistakes …

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12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance $1.21\ \text{Å}$ between them. The wavelength of the standing wave is (A) $1.21\ \text{Å}$ (B) $2.42\ \text{Å}$ (C) $6.05\ \text{Å}$ (D) $3.63\ \text{Å}$

    ›Reveal solutionSolution

    Three nodes (at both ends and the midpoint) with two antinodes is the second-harmonic pattern, for which the inter-atom separation equals exactly one wavelength.

    Concept and Intuition

    For a standing wave confined between two fixed points (nodes forced at both ends, as with the two atoms here), the general rule is: the $n$-th harmonic has $n$ antinodes and $n+1$ nodes, and the length $L$ between the fixed ends relates to the wavelength by $L = n\cdot\dfrac{\lambda}{2}$. Here we're told there are 2 antinodes and 3 nodes, which identifies $n=2$ (second harmonic).

    Step-by-Step Solution

    1. Number of antinodes given $=2 \Rightarrow n=2$ (this also gives $n+1=3$ nodes, matching the question — consistent).
    2. Relation between the confining length and wavelength: $L = n\cdot\dfrac{\lambda}{2} = 2\cdot\dfrac{\lambda}{2} = \lambda$. …

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