Two tuning forks of frequencies 250 and 256 Hz produce beats. If maximum is produced now, the time after which the minimum is produced at that same place is:

(1/18)s

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(1/24)s

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(1/6)s

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(1/12)s

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Solution

The correct option is C (1/12)s
$B e a t s = 256 - 250 = 6$
Time gap between maxima and minima is $= \frac{1}{b e a t s} \times \frac{1}{2}$
$t = \frac{1}{2 \times 6}$ $= \frac{1}{12}$ seconds

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Two tuning forks of frequencies 250 and 256 Hz produce beats. If maximum is produced now, the time after which the minimum is produced at that same place is:

The correct option is C (1/12)sBeats=256−250=6Time gap between maxima and minima is =1beats×12t=12×6 =112 seconds

The correct option is C (1/12)s
$B e a t s = 256 - 250 = 6$
Time gap between maxima and minima is $= \frac{1}{b e a t s} \times \frac{1}{2}$
$t = \frac{1}{2 \times 6}$ $= \frac{1}{12}$ seconds