A mass m is attached to four springs of spring constants 2.k,2k,-k and k as shown in fig. (2). The mass s capable of oscillating on friction less horizontal floor. If it is displaced slightly and released then the frequency of resulting S.H.M. would be
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a
12π11k2m
b
12π2k3m
c
12π3km
d
12π4km
answer is C.
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Detailed Solution
The resultant force constant is given by k′=(2k)(2k)2k+2k+(k+k)=3kWe know thatT=2πmk′=2πm3k Now n=1T=12π3km