The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

\sqrt{\frac{K}{m}}

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\sqrt{\frac{m}{K}}

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{(\frac{K}{m})}^{2}

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{(\frac{m}{K})}^{2}

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Solution

The correct option is A $\sqrt{\frac{K}{m}}$
Suppose you displace the particle by a distance $^{'} x^{'}$
The spring now exerts a force,
This provides nccenary force for $S H M$
$\Rightarrow F = m w e^{2} x = k 2$ ( $w :$ natural angular frequency )
$\Rightarrow w = \sqrt{K / m}$

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The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

The correct option is A √KmSuppose you displace the particle by a distance ′x′The spring now exerts a force,This provides nccenary force for SHM⇒ F=mwe2x=k2 (w: natural angular frequency )⇒ w=√K/m

The correct option is A $\sqrt{\frac{K}{m}}$
Suppose you displace the particle by a distance $^{'} x^{'}$
The spring now exerts a force,
This provides nccenary force for $S H M$
$\Rightarrow F = m w e^{2} x = k 2$ ( $w :$ natural angular frequency )
$\Rightarrow w = \sqrt{K / m}$