A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t1 If theperiod of oscillation with the two springs in series is T , then
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a
T=t1+t2
b
T2=t12+t22
c
Tâ1=t1â1+t2â1
d
Tâ2=t1â2+t2â2
answer is B.
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Detailed Solution
Here, t1=2Ïmk1 or t12=4Ï2mk1 and t2=2Ïmk2 or t22=4Ï2mk2 â´ t12+t22=4Ï2m1k1+1k3 When the springs are connected h seri6s, then 1k=1k1+1k2 Now T=2Ïmk =2Ïm1k1+1k2 or T2=4Ï2m1k1+1k2 T2=4Ï2m1k1+1k2 Fmm eqs. [1) atrd (2), we ge T2=t12+t22