JEE PHYSICS MCQ: WAVES

1 Medium

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Application +4 | -1

If a length of a closed organ pipe is

1\,\,m

and velocity of sound is

400\,\,m/s

, then the frequency for the second note is

View answer and explanation

Correct option: B

3 \times \frac{{330}}{4}Hz

Option B (

3 \times \frac{{330}}{4}Hz

) is correct because the second note in a closed organ pipe corresponds to the third harmonic, and the frequency is calculated using the formula

f_3 = \frac{{3v}}{{4l}}

.

More Questions from WAVES

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The source of sound generating a frequency of

3\,\,kHz

reaches an observer with a speed of

0.5

times, the velocity of sound in air. The frequency heard by the observer is A source of sound emits waves with frequency

f\mathrm{ }Hz

V\mathrm{ }m/sec

. Two observers move away from this source in opposite directions each with a speed

0.2\,\,V

relative to the source. The ratio of frequencies heard by the two observers will be Assertion: The wave function of a pulse is given by

y = \frac{3}{{{{\left( {2x + 3t} \right)}^2} + 4}},

it propagates in

(– \mathrm{ve}) \,x

-direction.
Reason: The given wave function is of the form

y = f(kx +\omega t),

which represent a wave travelling in negative

X

-direction. In the fundamental mode, time taken by the wave to reach the closed end of the air filled pipe is

0.01\,\,s

. the fundamental frequency is A rope

5 \mathrm{~m}

long has a total mass of

245 \mathrm{~g}

. It is stretched with a constant tension of

1 \mathrm{~kg}

wt. If it is fixed at one end and shaken by hand at the other end, what frequency of shaking will make it break up into three vibrating segments? Take

g=980 \mathrm{~cm} \mathrm{~s}^{-2}

T

is the reverberation time of an auditorium of volume

V