A wire having a linear density of 0.05 g/cm is stretched between two rigid supports with a tension of 4.5×102N it is observed that the wire resonance at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. Determine the length of the wire.
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a
2.14 m
b
3.2 m
c
4 m
d
8 m
answer is A.
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Detailed Solution
Expression for Frequency when both ends are tied f = n2LTμ Now, 420=n2LTμ→(i) 490=n+12LTμ→(ii) Divide equation i÷iin=6Putting in equation (i)420=62L4.5×1020.05×10−310−2⇒ L=2.14 m