JEE MAIN - Physics (2005 - No. 43)
When two tuning forks (fork $$1$$ and fork $$2$$) are sounded simultaneously, $$4$$ beats per second are heated. Now, some tape is attached on the prong of the fork $$2.$$ When the tuning forks are sounded again, $$6$$ beats per second are heard. If the frequency of fork $$1$$ is $$200$$ $$Hz$$, then what was the original frequency of fork $$2$$ ?
$$202$$ $$Hz$$
$$200$$ $$Hz$$
$$204$$ $$Hz$$
$$196$$ $$Hz$$
Explanation
No. of beats heard when fork $$2$$ is sounded with fork $$1$$ $$ = \Delta n = 4$$
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $$4$$ to $$6$$ in this case) then the frequency of the unknown fork $$2$$ is given by,
$$n = {n_0} - \Delta n = 200 - 4 = 196Hz$$
Now we know that if on loading (attaching tape) an unknown fork, the beat frequency increases (from $$4$$ to $$6$$ in this case) then the frequency of the unknown fork $$2$$ is given by,
$$n = {n_0} - \Delta n = 200 - 4 = 196Hz$$
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