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A particle at the end of a spring executes simple harmonic motion with a period ${t_1}$, while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then
$T = {t_1} + {t_2}$
${T^2} = t_1^2 + t_2^2$
${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$
${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$
Solution
(b) ${t_1} = 2\,\pi \sqrt {\frac{m}{{{k_1}}}} $ and ${t_2} = 2\,\pi \sqrt {\frac{m}{{{k_2}}}} $
In series, effective spring constant, $k = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}$
$\therefore $ Time period, $T = 2\,\pi \sqrt {\frac{{m\,({k_1} + {k_2})}}{{{k_1}{k_2}}}} $ …..(ii)
Now, $t_1^2 + t_2^2 = 4\,{\pi ^2}m\,\left( {\frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}} \right) = \frac{{4\,{\pi ^2}m\,({k_1} + {k_2})}}{{{k_1}{k_2}}}$
$t_1^2 + t_2^2 = {T^2}.$ [Using equation (ii)]
