Asked by Justin
0.460 kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to change the period to 1.95 s?
Answers
Answered by
Elena
T=2π•sqrt(m/k)
T₁²=4π²m₁/k
T₂²=4π²m₂/k
T₁²/T₂²=m₁/m₂
m₂=m₁T₂²/T₁²
Δm= m₂-m₁ =
=m₁(T₂²/T₁²-1)=
=0.46[(1.95/1.5)² -1] =0.317 kg
T₁²=4π²m₁/k
T₂²=4π²m₂/k
T₁²/T₂²=m₁/m₂
m₂=m₁T₂²/T₁²
Δm= m₂-m₁ =
=m₁(T₂²/T₁²-1)=
=0.46[(1.95/1.5)² -1] =0.317 kg
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