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BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Physics
Chapter: Waves and Sound

BITSAT 2009 · Physics · Waves and Sound

Two tuning forks A and B sounded together give 6 beats per second....

Two tuning forks

A

and

B

sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are

24 \mathrm{ cm}

and 25

\mathrm{cm}

respectively. Calculate the frequencies of forks.

Options

A $120 \mathrm{ Hz}, 124 \mathrm{ Hz}$
B $110 \mathrm{ Hz}, 114 \mathrm{ Hz}$
C $150 \mathrm{ Hz}, 144 \mathrm{ Hz}$
D $170 \mathrm{ Hz}, 118 \mathrm{ Hz}$

Correct answer

C. $150 \mathrm{ Hz}, 144 \mathrm{ Hz}$

Step-by-step solution

Let the frequency of the first fork be

\mathrm{f}_{1}

and that of second be

\mathrm{f}_{2}

We then have,

\mathrm{f}_{1}=\frac{\mathrm{v}}{4 \times 24}

and

\mathrm{f}_{2}=\frac{\mathrm{v}}{4 \times 25}

We also see that

\mathrm{f}_{1} > \mathrm{f}_{2} \therefore \mathrm{f}_{1}-\mathrm{f}_{2}=6 and \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{24}{25}

Solving (i) and (ii), we get

\mathrm{f}_{1}=150 \mathrm{ Hz}

and

\quad \mathrm{f}_{2}=144 \mathrm{ Hz}

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