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Question: A particle at the end of a spring executes simple harmonic motion with a period t<sub>1</sub>, while...

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 + t2t–1

D

T–2 = t1–2 + t2–2

Answer

T2 = t12 + t22

Explanation

Solution

t1 = 2p mk2\sqrt { \frac { \mathrm { m } } { \mathrm { k } _ { 2 } } }….(2)

When springs are in series then

T = 2p= 2p

squaring and adding (1) & (2) we get

+ = 4p2+4p2

= 4p2mor + = T2