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Question

A particle at the end of a spring executes simple harmonic motion 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:
  1. T=t1+t2
  2. T2=t21+t22
  3. T2=t21+t22
  4. T1=t11+t12

A
T2=t21+t22
B
T=t1+t2
C
T1=t11+t12
D
T2=t21+t22
Solution
Verified by Toppr

Let mass of the particle be m and k represents the spring constant.
Time period of oscillation of a particle executing SHM t=2πmk
Thus for 1st spring t1=2πmk1
k1=4π2mt21
Similarly k2=4π2mt22
Now spring constant of the combination of two springs in series 1kc=1k1+1k2
kc=k1k2k1+k2
Time period of the oscillation of combined springs T=2πmkc
kc=4π2mT2
k1k2k1+k2=4π2mT2

4π2mt21×4π2mt224π2mt21+4π2mt22=4π2mT2

1t21t221t21+1t22=1T2T2=t21+t22

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