A light pointer fixed to one prong of tuning fork touches a vertical plate.The fork is set vibrating and the plate is allowed to fall freely.8 oscillations are counted when plate falls through 10 cm.What is the frequency of tuning fork?

Plate falls through the dist. 0.1mtherfore time=underroot 2h by gThere fore we get time 1 by 7 seconds.Therefore in 1by 7 sec 8 oscillations occurs.Therefore in 1 sec 7×8= 56 oscillations made.As we knw frequency= no of cycles per second.=56HzI hope u understand. Good luck

Here we see that plate falls by a distance of 0.1m first we find in what time did plate take to fall 10cm . Same amount of time the fork would have taken to complete oscilattion .Required time is √2h/g. solving √0.02 s .Fork takes this time for 8 oscillation so then we know the time which it taje to complete on oscillation that is time period invese the value we will find frequency I.e. 40√2