A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes(54T). The ratio of mM is
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a
916
b
2516
c
45
d
54
answer is A.
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Detailed Solution
T = 2πmK ⇒m ∝T2 ⇒ m2m1 = T22T21⇒ M+mM = (54TT)2 ⇒ mM = 916