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Question

Physics Question on Oscillations

A mass is suspended separately by two different springs in successive order then time periods is t1t_1 and t2t_2 respectively. If it is connected by both spring as shown in figure then time period is t0t_0, the correct relation is

A

t02=t12+t22t_0^2 = t_1^2 + t_2^2

B

t02=t12+t22t_0^{ - 2} = t_1^{ -2} + t_2^{-2}

C

t01=t11+t21t_0^{ - 1} = t_1^{ - 1} + t_2^{ - 1}

D

t0=t1+t2t_0 = t_1 + t_2

Answer

t02=t12+t22t_0^{ - 2} = t_1^{ -2} + t_2^{-2}

Explanation

Solution

T=2πmK\because T =2 \pi \sqrt{\frac{ m }{ K }}
K1T2\Rightarrow K \propto \frac{1}{ T ^{2}}
In this case K=K1+K2K = K _{1}+ K _{2}
1t02=1t12+1t22\frac{1}{ t _{0}^{2}}=\frac{1}{ t _{1}^{2}}+\frac{1}{ t _{2}^{2}}
t02=t12+t22\Rightarrow t _{0}^{-2}= t _{1}^{-2}+ t _{2}^{-2}