When a mass of 0.5 kg is suspended from the free end of a spring, it stretches the spring by 0.2m. This mass is removed and 0.25kg mass is attached to the same free end of the spring. If the mass is pulled down and released, what is its time period?(g=10ms^(-2))

When a mass of 0.5 kg is suspended from the free end of a spring, it s

1.	When a mass of 0.5 kg is suspended from the free end of a spring, it stretches the spring by 0.2m. This mass is removed and 0.25kg mass is attached to the same free end of the spring. If the mass is pulled down and released, what is its time period?(g=10ms^(-2))
Answer» <html><body><p></p>Solution :When 0.3 kg <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> is suspended, extension of the spring is 0.2m. Since in equilibrium mg = Kx <br/> (or)`0.5 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>= k xx 0.2 <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> K= 25N//m` <br/> The time period of spring <a href="https://interviewquestions.tuteehub.com/tag/block-18865" style="font-weight:bold;" target="_blank" title="Click to know more about BLOCK">BLOCK</a> system is <br/> `T= 2pisqrt((m)/(k))= 2pisqrt((0.25)/(25))= 0.63 s`</body></html>