Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4$ beats ${{\rm{s}}^{ - 1}}$. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies are

$\begin{array}{*{20}{l}} {72\,\,and\,\,36} \end{array}$

$\begin{array}{*{20}{l}} {80\,\,and\,\,40} \end{array}$

$\begin{array}{*{20}{l}} {100\,\,and\,\,50} \end{array}$

$\begin{array}{*{20}{l}} {44\,\,and\,\,32} \end{array}$

JEE PHYSICS MCQ: WAVES

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Application +4 | -1

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce

4

beats

{\mathrm{s}^{ - 1}}

. The highest frequency is twice that of the lowest. Possible highest and lowest frequencies are

View answer and explanation

Correct option: D

\begin{array}{llllllllllllllllllll} {72\,\,and\,\,36} \end{array}

Option D (72 and 36) is correct because the highest frequency is twice the lowest, and the difference between adjacent forks is 4 Hz.

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3\,\,kHz

0.5

f\mathrm{ }Hz

V\mathrm{ }m/sec

0.2\,\,V

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

(– \mathrm{ve}) \,x

y = f(kx +\omega t),

X

0.01\,\,s

5 \mathrm{~m}

245 \mathrm{~g}

1 \mathrm{~kg}

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

T

V