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13.Oscillations
medium
A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ is
A
$\frac{1}{2 \pi} \sqrt{\frac{ K _1+ K _2}{ m }}$
B
$\frac{1}{2 \pi} \sqrt{\frac{ K _1- K _2}{ m }}$
C
$2 \pi \sqrt{\frac{ m }{ K _1+ K _2}}$
D
$2 \pi \sqrt{\frac{m}{K_1-K_2}}$
(JEE MAIN-2023)
Solution
On displacing $m$ to right by $x$
$F =-\left( k _1 x+ k _2 x \right)=-\left( k _1+ k _2\right) x$
$a =\frac{ F }{ m }=-\left(\frac{ k _1+ k _2}{ m }\right) x =-\omega^2 x$
$\therefore \omega=\sqrt{\frac{ k _1+ k _2}{ m }} \Rightarrow T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{ m }{ k _1+ k _2}}$
Standard 11
Physics
