JEE MAIN - Physics (2004 - No. 62)
A particle at the end of a spring executes $$S.H.M$$ 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
$${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$$
$${T^2} = t_1^2 + t_2^2$$
$$T = {t_1} + {t_2}$$
$${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$$
Explanation
For first spring, $${t_1} = 2\pi \sqrt {{m \over {{k_1}}}} ,$$
For second spring, $${t_2} = 2\pi \sqrt {{m \over {{k_2}}}} $$
when springs are in series then, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$
$$\therefore$$ $$T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2}}}} $$
$$\therefore$$ $$T = 2\pi \sqrt {{m \over {{k_2}}} + {m \over {{k_1}}}} $$
$$ = 2\pi \sqrt {{{t_2^2} \over {{{\left( {2\pi } \right)}^2}}} + {{t_1^2} \over {{{\left( {2\pi } \right)}^2}}}} $$
$$ \Rightarrow {T^2} = t_1^2 + t_2^2$$
For second spring, $${t_2} = 2\pi \sqrt {{m \over {{k_2}}}} $$
when springs are in series then, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$
$$\therefore$$ $$T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2}}}} $$
$$\therefore$$ $$T = 2\pi \sqrt {{m \over {{k_2}}} + {m \over {{k_1}}}} $$
$$ = 2\pi \sqrt {{{t_2^2} \over {{{\left( {2\pi } \right)}^2}}} + {{t_1^2} \over {{{\left( {2\pi } \right)}^2}}}} $$
$$ \Rightarrow {T^2} = t_1^2 + t_2^2$$
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