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प्रश्न
A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire Bproduces 6 beats per second with it when the tension in A is slightly increased. Find the the ratio of the tension in A to the tension in B.
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उत्तर
Mass per unit length of both the wires = m
Fundamental frequency of wire of length \[\left( l \right)\] and tension \[\left( T \right)\] is given by :
\[n = \frac{1}{2I}\sqrt{\frac{T}{m}}\]
It is clear from the above relation that as the tension increases, the frequency increases.
Fundamental frequency of wire A is given by : \[n_A = \frac{1}{2I}\sqrt{\frac{T_A}{m}}\]
Fundamental frequency of wire B is given by:
\[n_B = \frac{1}{2I}\sqrt{\frac{T_B}{m}}\]
It is given that 6 beats are produced when the tension in A is increased.
⇒ \[n_A = 606 = \frac{1}{2l}\sqrt{\frac{T_A}{m}}\]
Therefore, the ratio can be obtained as:
\[ \frac{n_A}{n_B} = \frac{606}{600} = \frac{\left( 1/2I \right)\sqrt{\left( T_A /m \right)}}{\left( 1/2I \right)\sqrt{T_B /m}}\]
\[ \Rightarrow \frac{606}{600} = \frac{\sqrt{T_A}}{\sqrt{T_B}}\]
\[ \Rightarrow \frac{\sqrt{T_A}}{\sqrt{T_B}} = \frac{606}{600} = 1 . 01\]
\[ \Rightarrow \frac{T_A}{T_B} = 1 . 02\]
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