When a body of mass $$1.0$$ kg is suspended from a certain light spring hanging vertically, its length increases by $$5$$ cm. By suspending $$2$$ $$.0$$ kg block to the spring and if the block is pulled through $$10$$ cm and released, the maximum velocity in it in $$m/s$$ is [Acceleration due to gravity $$=$$ $$10$$ m $$/$$ $$s2)$$

Question

Infinity Learn · Accepted Answer

For a spring F $$=-k$$ x and F $$=m$$ gm g $$=$$ $$-$$ k x or $$k=(mg/x)$$ magnitude  or $$k=1$$×$$10(5/100)=200N/m$$ The angular frequency of oscillation of $$2$$ kg mass is given by ω$$=km=2002=10rad/$$$$sec$$  Now $$vmax=$$ω$$a=10$$×$$0$$⋅$$1=1m/s$$

When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2 .0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity in it in m/s is [Acceleration due to gravity = 10 m / s²)

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When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2 .0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity in it in m/s is [Acceleration due to gravity = 10 m / s2)

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When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending 2 .0 kg block to the spring and if the block is pulled through 10 cm and released, the maximum velocity in it in m/s is [Acceleration due to gravity = 10 m / s²)