JEE Advance - Physics (2009 - Paper 2 Offline - No. 3)
The mass M shown in the figure below oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is
$${{{k_1}A} \over {{k_2}}}$$
$${{{k_2}A} \over {{k_1}}}$$
$${{{k_1}A} \over {{k_1} + {k_2}}}$$
$${{{k_2}A} \over {{k_1} + {k_2}}}$$
Explanation
Since the restoring force is same in both springs (which are being in series), we have
$${k_1}{x_1} = {k_2}{x_2}$$
It is given that
$${x_1} + {x_2} = A$$
$$ \Rightarrow {x_1} = {{A{k_2}} \over {{k_1} + {k_2}}}$$
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