Question

# A particle at the end of a spring executes simple harmonic motion with a period $$\mathrm{t}_{1}$$, while the corresponding period for another spring is $$\mathrm{t}_{2}$$. If the period of oscillation with the two springs in series is $$\mathrm{T}$$, then:

A
T=t1+t2
B
T2=t21+t22
C
T1=t11+t12
D
T2=t21+t22

Solution

## The correct option is B $$\mathrm{T}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}$$Let mass of the particle be $$m$$   and   $$k$$ represents  the spring constant.Time period of oscillation of a particle executing SHM         $$t = 2\pi \sqrt{\dfrac{m}{k}}$$Thus for 1st spring       $$t_1 = 2\pi \sqrt{\dfrac{m}{k_1}}$$  $$\implies k_1 = \dfrac{4\pi^2 m}{t_1^2}$$Similarly        $$k_2 = \dfrac{4\pi^2 m}{t_2^2}$$Now spring constant of the combination of two springs in  series       $$\dfrac{1}{k_c} = \dfrac{1}{k_1} + \dfrac{1}{k_2}$$$$\implies k_c = \dfrac{k_1 k_2}{k_1+k_2}$$Time period of the oscillation of combined springs          $$T = 2\pi \sqrt{\dfrac{m}{k_c}}$$$$\therefore$$  $$k_c = \dfrac{4\pi^2 m}{T^2}$$$$\dfrac{k_1 k_2}{k_1+k_2} = \dfrac{4\pi^2 m}{T^2}$$$$\dfrac{\dfrac{4\pi^2 m}{t_1^2} \times \dfrac{4\pi^2 m}{t_2^2}}{\dfrac{4\pi^2 m}{t_1^2} + \dfrac{4\pi^2 m }{t_2^2}}= \dfrac{4\pi^2 m}{T^2}$$$$\dfrac{\dfrac{1}{t_1^2 t_2^2}}{\dfrac{1}{t_1^2} + \dfrac{1}{t_2^2} } = \dfrac{1}{T^2} \implies T^2 = t_1^2 + t_2^2$$Physics

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