Determination of Unknown Frequency
Category : JEE Main & Advanced
Suppose a tuning fork of known frequency \[({{n}_{A}})\] is sounded together with another tuning fork of unknown frequency \[({{n}_{B}})\] and \[x\] beats heard per second.
There are two possibilities to known frequency of unknown tuning fork.
\[{{n}_{A}}-{{n}_{B}}=x\] ... (i)
or \[{{n}_{B}}-{{n}_{A}}=x\] ... (ii)
To find the frequency of unknown tuning fork \[({{n}_{B}})\] following steps are taken.
(1) Loading or filing of one prong of known or unknown (by wax) tuning fork, so frequency changes (decreases after loading, increases after filing).
(2) Sound them together again, and count the number of heard beats per sec again, let it be \[x'\]. These are following four condition arises.
(i) \[x'>x\]
(ii) \[x'<x\]
(iii) \[x'=0\]
(iv) \[x'=x\]
(3) With the above information, the exact frequency of the unknown tuning fork can be determined as illustrated below.
Suppose two tuning forks A (frequency \[{{n}_{A}}\] is known) and B (frequency \[{{n}_{B}}\] is unknown) are sounded together and gives \[x\] beats/sec. If one prong of unknown tuning fork B is loaded with a little wax (so \[{{n}_{B}}\] decreases) and it is sounded again together with known tuning fork A, then in the following four given condition \[{{n}_{B}}\] can be determined.
(4) If \[x'>x\] than \[x,\] then this would happen only when the new frequency of B is more away from \[{{n}_{A}}\]. This would happen if originally (before loading), \[{{n}_{B}}\] was less than \[{{n}_{A}}\].
Thus initially \[{{n}_{B}}={{n}_{A}}-x\].
(5) If \[x'<x\] than \[x,\] then this would happen only when the new frequency of B is more nearer to \[{{n}_{A}}\]. This would happen if originally (before loading), \[{{n}_{B}}\] was more than \[{{n}_{A}}\]. Thus initially \[{{n}_{B}}={{n}_{A}}+x\].
(6) If \[x'=x\] then this would means that the new frequency (after loading) differs from \[{{n}_{A}}\] by the same amount as was the old frequency (before loading). This means initially \[{{n}_{B}}={{n}_{A}}+x\]
(and now it has decreased to \[n{{'}_{B}}={{n}_{A}}-x\])
(7) If \[x'=0,\] then this would happen only when the new frequency of B becomes equal to \[{{n}_{A}}\] This would happen if originally \[{{n}_{B}}\] was more than \[{{n}_{A}}\].
Thus initially \[{{n}_{B}}=n+x\].
Frequency of unknown tuning fork for various cases
| By loading | |
| If B is loaded with wax so its frequency decreases | If A is loaded with wax its frequency decreases |
| If \[x\] increases \[{{n}_{B}}={{n}_{A}}-x\] | If \[x\] increases \[{{n}_{B}}={{n}_{A}}+x\] |
| If \[x\] decrease \[{{n}_{B}}={{n}_{A}}+x\] | If \[x\] decrease \[{{n}_{B}}={{n}_{A}}-x\] |
| If remains same \[{{n}_{B}}={{n}_{A}}+x\] | If remains same \[{{n}_{B}}={{n}_{A}}-x\] |
| If \[x\] becomes zero \[{{n}_{B}}={{n}_{A}}+x\] | If \[x\] becomes zero \[{{n}_{B}}={{n}_{A}}-x\] |
| By filing | |
| If B is filed, its frequency increases | If A is filed, its frequency increases |
| If \[x\] increases \[{{n}_{B}}={{n}_{A}}+x\] | If \[x\] increases \[{{n}_{B}}={{n}_{A}}-x\] |
| If \[x\] decrease\[{{n}_{B}}={{n}_{A}}-x\] | If \[x\] decrease \[{{n}_{B}}={{n}_{A}}+x\] |
| If remains same \[{{n}_{B}}={{n}_{A}}-x\] | If remains same \[{{n}_{B}}={{n}_{A}}+x\] |
| If \[x\] becomes zero \[{{n}_{B}}={{n}_{A}}-x\] | If \[x\] becomes zero \[{{n}_{B}}={{n}_{A}}+x\] |
You need to login to perform this action.
You will be redirected in
3 sec
