Wave on String – JEE Main Previous Year Questions with Solutions
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Q. The equation of a wave on a string of linear mass density 0.04 kg $\mathrm{m}^{-1}$ is given by y $=0.02(\mathrm{m}) \sin \left[2 \pi\left(\frac{\mathrm{t}}{0.04(\mathrm{s})}-\frac{\mathrm{x}}{0.50(\mathrm{m})}\right)\right]$. The tension in the string is :(1) 6.25 N           (2) 4.0 N              (3) 12.5 N            (4) 0.5 N [AIEEE – 2010]

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Sol. (1)$y=0.2 \sin \left[2 \pi\left(\frac{t}{0.04}-\frac{x}{0.50}\right)\right]$$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{\omega}{\mathrm{k}} \Rightarrow \sqrt{\frac{\mathrm{T}}{0.04}}=\frac{\frac{1}{0.04}}{\frac{1}{0.50}}$$\mathrm{T}=\left(\frac{0.50}{0.04}\right)^{2} \times 0.04=6.25 \mathrm{N}$

Q. The transverse displacement y(x, t) of a wave on a string is given by $y(\mathrm{x}, \mathrm{t})=e^{-\left(\mathrm{ax}^{2}+\mathrm{bt}^{2}+2 \sqrt{\mathrm{ab}} \mathrm{xt}\right)}$. This represents a :-(1) standing wave of frequency $\sqrt{b}$(2) standing wave of frequency $\frac{1}{\sqrt{\mathrm{b}}}$(3) wave moving in $+x$ directionwith speed $\sqrt{\frac{a}{b}}$(4) wave moving in -x direction with speed $\sqrt{\frac{b}{a}}$ [AIEEE – 2011]

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Sol. (4)$y\left(x_{1} t\right)=e^{-[\sqrt{a x}+\sqrt{b t}]^{2}}$$\mathrm{v}=\omega / \mathrm{K}=\frac{\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}} in -ve \mathrm{x} direction. Q. Statement-1: Two longitudinal waves given by equations : \mathrm{y}_{1}(\mathrm{x}, \mathrm{t})=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}) and \mathrm{y}_{2} (x, t) = a sin (2\omegat – 2kx) will have equal intensity.Statement-1: Intensity of waves of given frequency in same medium is proportional to square of amplitude only.(1) Statement-1 is false, statement-2 is true(2) Statement-1 is ture, statement-2 is false(3) Statement-1 is ture, statement-2 true; statement-2 is the correct explanation of statement-1(4) Statement-1 is true, statement-2 is true; statement -2 is not correct explanation of statement-1. [AIEEE – 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (2)\mathrm{y}_{1}(\mathrm{x}, \mathrm{t})=2 \mathrm{a} \sin (\mathrm{wt}-\mathrm{kx})$$y_{2}(\mathrm{x}, \mathrm{t})=\mathrm{a} \sin (2 \mathrm{wt}-2 \mathrm{kx})$But Intensity $\mathrm{I}=2 \pi^{2} n^{2} \mathrm{a}^{2} \rho \mathrm{v} \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\left(\frac{2 \mathrm{a}}{\mathrm{a}} \times \frac{\mathrm{n}}{2 \mathrm{n}}\right)^{2}=\frac{1}{1}$Intensity depends on frequency and amplitudeSo statement- 1 is true statement- 2 is false

Q. A travelling wave represented by y = A sin($\omega$t–kx) is superimposed on another wave represented by y = A sin ($\omega$t + kx). The resultant is :-(1) A standing wave having nodes at $x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}, n=0,1,2$(2) A wave travelling along $+x$ direction(3) A wave travelling along – x direction(4) A standing wave having nodes at $\mathrm{x}=\frac{\mathrm{n} \lambda}{2} ; \mathrm{n}=0,1,2$ [AIEEE-2011]

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Sol. (1)$y_{1}=A \sin (w t-k x) \& y_{2}=A \sin (w t+k x)$By superposition principle$y=y_{1}+y_{2}$$=A \sin (w t-k x)+A \sin (w t+k x)$$=2 \mathrm{A} \sin \mathrm{wt} \cos \mathrm{kx}$Amplitude = 2A cos kxAt nodes displacement is minimum$2 \mathrm{A} \cos \mathrm{kx}=0 \Rightarrow \cos \mathrm{kx}=0$$\mathrm{kx}=(2 \mathrm{n}+1) \frac{\pi}{2} \Rightarrow \frac{2 \pi}{2} \mathrm{x}=(2 \mathrm{n}+1) \frac{\pi}{2}$$\mathrm{x}=(2 \mathrm{n}+1) \frac{\pi}{4}$ where $\mathrm{n}=0,1,2 \ldots$

Q. A sonometer wire of length 1.5m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × $10^{3}$ $\mathrm{kg} / \mathrm{m}^{3}$ and 2.2 × $10^{11} \mathrm{N} / \mathrm{m}^{2}$ respectively ?(1) 188.5 Hz            (2) 178.2 Hz            (3) 200.5 Hz              (4) 770 Hz [JEE-Main-2013]

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Sol. (2)Fundamental frequency $\mathrm{f}=\frac{\mathrm{V}}{2 \ell}=\frac{1}{2 \times 1.5} \sqrt{\frac{\mathrm{T}}{\mathrm{eq}}}=\frac{1}{3} \sqrt{\frac{\mathrm{y} \times \operatorname{strain} \times \mathrm{S}}{\rho \mathrm{S}}}$(S  cross – section Area)$=\frac{1}{3} \sqrt{\frac{2.2 \times 10^{11} \times \frac{1}{100}}{7.7 \times 10^{3}}}=178.2 \mathrm{Hz}$

Q. A uniform string of length 20m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is :-(take $\left.\mathrm{g}=10 \mathrm{ms}^{-2}\right)$(1) $\sqrt{2} \mathrm{s}$(2) $2 \pi \sqrt{2} \mathrm{s}$(3) 2s(4) $2 \sqrt{2} \mathrm{s}$ [JEE-Main-2016]

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Sol. (4)

Q. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequencty of the air column is now :-(1) f(2) $\frac{\mathrm{f}}{2}$(3) $\frac{3 \mathrm{f}}{4}$(4) 2f [JEE-Main-2016]

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Sol. (1)

• January 10, 2022 at 12:00 am

Sir
Sare questions ban gye sir
Thanks 😊 sir

• March 1, 2021 at 10:29 pm

Upload more questions but in Hindi

• February 8, 2021 at 3:59 pm

• September 29, 2020 at 11:48 pm

Need more questions from recent years

• August 28, 2020 at 6:43 pm

• August 26, 2020 at 10:53 pm

Thanks man 👍👍 Its really helpful in boosting up the confidence

• July 26, 2020 at 11:07 pm

Weather is so COOL!!!!!!!!!!!!
But Problems are too HOT……….

• July 26, 2020 at 11:04 pm

excellent

• July 14, 2020 at 11:08 pm