A spring of force constant 180 N/m is loaded with a mass 0.20 kg. The amplitude of oscillations is 4 cm. When mass comes to equillibrium position, its velocity in m/s is

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The angular frequency of the spring-mass system is given by $ω = \frac{k}{m}$ .

Substituting the given values, $k = 180$ N/m and $m = 0.20$ kg:

$ω = \frac{180}{0.20} = 900 = 30$ rad/s.

The velocity of the mass at the equilibrium position is its maximum velocity, which is given by $v = A ω$ .

Given the amplitude $A = 4$ cm $= 0.04$ m:

$v = 0.04 \times 30 = 1.2$ m/s.

Answer: $1.2$

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