Q141JEE Main 2006MCQ

A whistle producing sound waves of frequencies $9500 {\text{Hz}}$ and above is approaching a stationary person with speed $v {\text{ms}}^{-1}$ . The velocity of sound in air is $300 {\text{ms}}^{-1}$ . If the person can hear frequencies upto a maximum of $10,000 {\text{HZ}}$ , the maximum value of $v$ upto which he can hear whistle is

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📖 Explanation

$v^{'}=v [\;\frac{v}{v-v_s}] \Rightarrow 10000$ $=9500 [\;\frac{300}{300-v}]$ $\Rightarrow 300-v=300 \times 0.95 \Rightarrow v$ $=300-285=15 {\text{ms}}^{-1}$

Q142JEE Main 2005MCQ

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

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📖 Explanation

$n^{'}=n [\;\frac{v+v_0}{v}] =n [\;\frac{v+\frac{v}{5}}{v}] =n [\;\frac{6}{5}]$ $\;\frac{n^{'}}{n}=\;\frac{6}{5} ; \;\frac{n^{'}-n}{n}$ $=\;\frac{6-5}{5} \times 100=20 \%$

Q143JEE Main 2005MCQ

When two tuning forks (fork 1 and fork 2 ) are sounded simultaneously, 4 beats per second are heated. Now, some tape is attached on the prong of the fork 2 . When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is $200 {\text{Hz}}$ , then what was the original frequency of fork 2 ?

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📖 Explanation

No. of beats heard when fork 2 is sounded with fork $1=\Delta n=4$ Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from 4 to 6 in this case) then the frequency of the unknown fork 2 is given by, $n=n_0-\Delta n=200-4=196 {\text{Hz}}$

Q144JEE Main 2004MCQ

The displacement $y$ of a particle in a medium can be expressed as, $y=10^{-6} {\ sin}$ $(100 t+20 x+\;\frac{\pi }{4}) m$ where $t$ is in second and $x$ in meter. The speed of the wave is

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📖 Explanation

From equation given, $\omega =100$ and $k=20, v=\;\frac{\omega }{k}=\;\frac{100}{20}=5 \frac{\text{m}}{\ s}$

Q145JEE Main 2003MCQ

A tuning fork of known frequency $256 {\text{Hz}}$ makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

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📖 Explanation

A tuning fork of frequency $256 {\text{Hz}}$ makes 5 beats/ second with the vibrating string of a piano.Therefore the frequency of the vibrating string of piano is $(256 ± 5) {\text{Hz}}$ ie either $261 {\text{Hz}}$ or $251 {\text{Hz}}$ . When the tension in the piano string increases, its frequency will increases. Now since the beat frequency decreases, we can conclude that the frequency of piano string is $251 {\text{Hz}}$

Q146JEE Main 2003MCQ

A metal wire of linear mass density of $9.8 \frac{\text{g}}{\text{m}}$ is stretched with a tension of $10 {\text{kg}}-w {\text{t}}$ between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $n$ . The frequency $n$ of the alternating source is

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📖 Explanation

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by $n=\;\frac{1}{2 ℓ} \sqrt{\;\frac{T}{\mu }}=\;\frac{1}{2 \times } \sqrt{\;\frac{10 \times 9.8}{9.8 \times 10^{-3}}}=50 H z$ As the string is vibrating in resonance to a.c of frequency $n$ , therefore both the frequencies are same.

Q147JEE Main 2003MCQ

The displacement $y$ of a wave travelling in the $x$ -direction is given by $y=10^{-4} sin \; (600 t-2 x+\;\frac{\pi }{3}) \;\text{ metres }\;$ where $x$ is expressed in metres and $t$ in seconds. The speed of the wave - motion, in $m s^{-1}$ , is

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📖 Explanation

$y=10^{-4} sin \; (600 t-2 x+\;\frac{\pi }{3})$ But $y=A sin \;(\omega t-k x+\varphi )$ On comparing we get $\omega =600 ; k=2$ $v=\;\frac{\omega }{k}=\;\frac{600}{2}=300 {\text{ms}}^{-1}$

Q148JEE Main 2002MCQ

A wave $y=a \sin (\omega t-k x)$ on a string meets with another wave producing a node at $x=0$ . Then the equation of the unknown wave is

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📖 Explanation

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $\pi$ . The equation of the reflected wave will be $y=a \sin (\omega t+k x+\pi )$ $\Rightarrow y=-a \sin (\omega t+k x)$

Q149JEE Main 2002MCQ

Tube $A$ has bolt ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is

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📖 Explanation

KEY CONCEPT : The fundamental frequency for closed organ pipe is given by $v_c=\;\frac{v}{4 ℓ}$ and For open organ pipe is given by $v_0=\;\frac{v}{2 ℓ}$ $\therefore \;\frac{v_0}{v_c}=\;\frac{v}{2 ℓ} \times \;\frac{4 ℓ}{v}=\;\frac{2}{1}$

Q150JEE Main 2002MCQ

length of a string tied to two rigid supports is $40 {\text{cm}}$ . Maximum length (wavelength in ${\text{cm}}$ ) of a stationary wave produced on it is

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📖 Explanation

This will happen for fundamental mode of vibration as shown in the figure. $S_1$ and $S_2$ are rigid support Here $\;\frac{\lambda }{2}=40 \;\; \therefore \lambda =80 {\text{cm}}$

Q151JEE Main 2002MCQ

A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency $288 {\text{cps}}$ . A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is

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📖 Explanation

A tuning fork produces 4 beats/sec with another tuning fork of frequency $288 {\text{cps}}$ . From this information we can conclude that the frequency of unknown fork is $288+4 {\text{cps}}$ or $288-4 {\text{cps}}$ i.e. $292 {\text{cps}}$ or $284 {\text{cps}}$ . Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4 . So the difference between frequency decreases. This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork. So the frequency of the unknown tuninh fork is $=292 {\text{Hz}}$

Q152JEE Main 2002MCQ

When temperature increases, the frequency of a tuning fork

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📖 Explanation

KEY CONCEPT : The frequency of a tuning fork is given by the expression $f=\;\frac{m^2 k}{4 \sqrt{3} \pi ℓ^2} \sqrt{\;\frac{Y}{\rho }}$ As temperature increases, $ℓ$ increases and therefore $f$ decreases.