The system shown lies on a horizontal smooth surface. The mass ' $m$ ' is given a small displacement along the x - axis. The time period will be:

2 π \sqrt{\frac{2 m}{5 k}}

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2 π \sqrt{\frac{5 m}{2 k}}

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2 π \sqrt{\frac{3 m}{2 k}}

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None

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Solution

The correct option is A $2 π \sqrt{\frac{2 m}{5 k}}$

$x^{'} = x c o s 60^{\circ} = \frac{x}{2}$

restoring force $(F_{R}) :$
$F_{R} = - [\frac{k x}{2} cos 60^{\circ} + \frac{k x}{2} cos 60^{\circ} + 2 k x]$
$F_{R} = - \frac{5 k x}{2}$
Now, time - period
$T = 2 π \sqrt{\frac{m}{k_{e q}}}$
where $k_{e q}$ is the equivalent spring constant
$T = 2 π \sqrt{\frac{2 m}{5 k}}$
Hence, option (A) is correct.

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The system shown lies on a horizontal smooth surface. The mass 'm' is given a small displacement along the x - axis. The time period will be:

The correct option is A 2π√2m5k x′=x cos 60∘=x2 restoring force (FR): FR=−[kx2cos60∘+kx2cos60∘+2kx] FR=−5kx2 Now, time - period T=2π√mkeq where keq is the equivalent spring constant T=2π√2m5k Hence, option (A) is correct.

The system shown lies on a horizontal smooth surface. The mass ' $m$ ' is given a small displacement along the x - axis. The time period will be: