Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On

sounding these forks together, the time interval between successive

maximum intensities will be

[MP PET 1989; MP PMT 2003]

sec

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1 sec

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Solution

The correct option is A
sec

The time interval between successive maximum intensities will be

$\frac{1}{n_{1} \sim n_{2}} = \frac{1}{454 - 450} = \frac{1}{4}$ sec

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Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On sounding these forks together, the time interval between successive maximum intensities will be [MP PET 1989; MP PMT 2003]

The correct option is A sec The time interval between successive maximum intensities will be 1n1∼n2=1454−450=14 sec

Two tuning forks have frequencies 450 Hz and 454 Hz respectively. On

sounding these forks together, the time interval between successive

maximum intensities will be

[MP PET 1989; MP PMT 2003]

The correct option is A
sec

The time interval between successive maximum intensities will be

$\frac{1}{n_{1} \sim n_{2}} = \frac{1}{454 - 450} = \frac{1}{4}$ sec