Two tuning forks of frequencies 250 Hz and 256 Hz produce beats. If a maximum is produced just now, after how much time the minima is produced at the same plane -

\frac{1}{18} s e c

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\frac{1}{24} s e c

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\frac{1}{6} s e c

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\frac{1}{12} s e c

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Solution

The correct option is D $\frac{1}{12} s e c$
Beat frequency, $f_{b} = | f_{2} - f_{1} | = | 256 - 250 | = 6 H z$

Time from one maxima to minima

$T^{'} = \frac{T_{b}}{2} = \frac{1}{2 f_{b}} = \frac{1}{12} s e c$

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Two tuning forks of frequencies 250 Hz and 256 Hz produce beats. If a maximum is produced just now, after how much time the minima is produced at the same plane -

The correct option is D 112secBeat frequency, fb=|f2−f1|=|256−250|=6HzTime from one maxima to minimaT′=Tb2=12fb=112sec

The correct option is D $\frac{1}{12} s e c$
Beat frequency, $f_{b} = | f_{2} - f_{1} | = | 256 - 250 | = 6 H z$

Time from one maxima to minima

$T^{'} = \frac{T_{b}}{2} = \frac{1}{2 f_{b}} = \frac{1}{12} s e c$