A body at the end of a spring executes S.H.M. with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then
A
T=t1+t2
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B
T2=t21+t22
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C
1T=1t1+1t2
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D
1T2=1t21+1t22
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Solution
The correct option is BT2=t21+t22 Let mass of the particle be m and k represents the spring constant.
Time period of oscillation of a particle executing SHM t=2π√mk
Thus, for first spring t1=2π√mk1 ⇒k1=4π2nt21
Now spring constant of the combination of two springs series 1kc=1k1+1k2
Time period of the oscillation of combined springs T=2π√mkc ∴kc=4π2mT2 k1k2k1+k2=4π2mT2 4π2mt21×4π2mt224π2mt21+4π2mt22=4π2mT2 1t21×t221t21+1t22=1T2⇒T2=t21+t22