A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of ‘m’ are attached at distance ‘L/2’ from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to:
(A) 0.77 (B) 0.57
(C) 0.37 (D) 0.17
Answers
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Answer:
C. will be correct answers
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Thus the ratio between the masses is m / M = 0.375
Option (C) is correct.
Explanation:
Frequency of torsonal oscillations is given by :
f = k / √I
f1 = k / √M (2L)^2 / 12
f2 = k / √ M(2L)^2 / 12 + 2m (L/2)^2
f2 = 0.8 f1
m / M = 0.375
Thus the ratio between the masses is m / M = 0.375
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