QuestionDetails: If the clock losses 10 min/day when the length of the pend is30

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QuestionDetails: If the clock losses 10 min/day when the length of the pend is30 inches. With what length pendulum will the pend keepperfect time? Equations that may help:    g1/g2=R22/ R12    Period 2/period 1 = R2/R1(l2/ l1)1/2 R is radius and l is length g is usually 32 You can try it without these equations ....anyway willdo QuestionDetails: If the clock losses 10 min/day when the length of the pend is30 inches. With what length pendulum will the pend keepperfect time? Equations that may help:    g1/g2=R22/ R12    Period 2/period 1 = R2/R1(l2/ l1)1/2 R is radius and l is length g is usually 32 You can try it without these equations ....anyway willdo R is radius and l is length g is usually 32 You can try it without these equations ....anyway willdo

Explanation / Answer

if the pendulum loses 10 minutes per day then "n" cyclestakes   1430 minutes (one day minus 10minutes). . You want a pendulum that completes n cyclesin   1440 minutes (one day exactly). . So...     1430 = nT1       1440 = nT2 . The period of a pendulum is given by    T = 2 L/g     so... .         1430 = n 2 L1 /g           1440= n 2 L2 / g . If you take the ratio of the equations, you can seethat    n 2   and g allcancel on the right side, so... .         1440 /1430   = L2 /L1 . square both sides:     (1440/1430)2 = L2 / 30 inches .    L2 = 30 inches * 1.014035   =    30.421 inches is the length of the pendulum that keeps correct time . The period of a pendulum is given by    T = 2 L/g     so... .         1430 = n 2 L1 /g           1440= n 2 L2 / g . If you take the ratio of the equations, you can seethat    n 2   and g allcancel on the right side, so... .         1440 /1430   = L2 /L1 . square both sides:     (1440/1430)2 = L2 / 30 inches .    L2 = 30 inches * 1.014035   =    30.421 inches is the length of the pendulum that keeps correct time

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