JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 16)
5 beats / econd are heard when a tuning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be :
195 Hz
150 Hz
300 Hz
251 Hz
Explanation
Length of wire is L1 = 0.95 m; L2 = 1 m. Number of beats per second heard = 5
Let frequency of fork be f. Therefore,
$${v \over {2{L_1}}} - f = 5 \Rightarrow {v \over {2{L_1}}} = 5 + f$$ ...... (1)
and $$f - {v \over {2{L_2}}} = 5 \Rightarrow {v \over {2{L_2}}} = f - 5$$ .... (2)
Dividing Eq. (1) by Eq. (2), we get
$${{{v \over {2{L_1}}}} \over {{v \over {2{L_2}}}}} = {{5 + f} \over {f - 5}} \Rightarrow {{{L_2}} \over {{L_1}}} = {{f + 5} \over {f - 5}}$$
$$ \Rightarrow {1 \over {0.95}} = {{f + 5} \over {f - 5}} \Rightarrow f - 5 = 0.95f + 4.75$$
$$ \Rightarrow f - 0.95f = 5 + 4.75 \Rightarrow 0.05f = 9.75 \Rightarrow f = {{9.75} \over {0.05}}$$
$$ \Rightarrow f = 195$$ Hz
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