JEE MAIN - Physics (2002 - No. 43)
A tuning fork arrangement (pair) produces $$4$$ beats/sec with one fork of frequency $$288$$ $$cps.$$ A little wax is placed on the unknown fork and it then produces $$2$$ beats/sec. The frequency of the unknown fork is
$$286$$ $$cps$$
$$292$$ $$cps$$
$$294$$ $$cps$$
$$288$$ $$cps$$
Uitleg
A tuning fork produces $$4$$ beats/sec with another tuning fork of frequency $$288$$ cps. From this information we can conclude that the frequency of unknown fork is $$288+4$$ $$cps$$ or $$288-4$$ $$cps$$ i.e. $$292$$ $$cps$$ or $$284$$ $$cps.$$
Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.
This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.
So the frequency of the unknown tuninh fork is = 292 Hz
Here when a little wax is placed on the unknown fork, it decreases the frequency of unknown fork. Here also beats per second decreases to 2 from 4. So the difference between frequency decreases.
This is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of the given tuning fork.
So the frequency of the unknown tuninh fork is = 292 Hz