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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
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B
T2=t21+t22
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C
T1=t11+t12
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D
T2=t21+t22
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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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