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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

a
T=t1+t2
b
T2=t12+t22
c
T−1=t1−1+t2−1
d
T−2=t1−2+t2−2

detailed solution

Correct option is B

t1=2 πmk1 and  t2=2 πmk2In series, effective spring constant,  k=k1k2k1+k2∴Time period, T=2 πm (k1+k2)k1k2     …..(ii) Now,  t12+t22=4 π2m 1k1+1k2=4 π2m (k1+k2)k1k2t12+t22=T2.                      [Using equation (ii)]

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