Question

The scale of a spring balance reading from 0 to 10kg is 0.25m long. A body suspended from the balance oscillates vertically with a period of /10s. The mass suspended is neglected then the mass of the spring is:

Answer 1

Answer:

Explanation:

Using F=kx ⇒10g=k×0.25⇒k=10g0.25=98×4                

   Now T=2πmk−−−√⇒m=T24π2k          

 ⇒m=π2100×14π2×98×4=0.98 kg