When a mass of $$0.5$$ kg is suspended from the free end of spring it stretches the spring by $$0.2$$ m. This mass is removed and $$0.25$$ kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period?(g=10ms-1)

Question

Infinity Learn · Accepted Answer

When $$0.5$$ kg is suspended, extension of the spring is $$0.2$$ m. Since in equilibrium $$mg=kxWhen$$ $$0.5$$ kg is suspended, extension of the spring is $$0.2$$ m.  in equilibrium $$mg=$$$$kx$$ $$0.5$$×$$10=k$$×$$0.2$$ spring constant $$=k=25$$ $$N/m$$  Time period of spring block system when $$0.25$$ kg mass is attached time period $$=T=2$$$$π$$mK substitute m, K values $$T=2$$$$π$$$$0.2525$$ $$T=2$$$$π$$$$252500$$ $$T=2$$×$$3.1410$$ $$T=0.628$$ seconds

When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period? $(g = 10 m s^{- 1})$

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When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period?(g=10ms-1)

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When a mass of 0.5 kg is suspended from the free end of spring it stretches the spring by 0.2 m. This mass is removed and 0.25 kg mass is attached to the same free end of the spring. If the mass is pulled down and released. What is it’s time period? $(g = 10 m s^{- 1})$