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Author Question: A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It ... (Read 67 times)

rosent76

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on: Aug 6, 2020
A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately

◦
y = −0.10 sin 9.9t

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y = 0.10 cos 9.9t

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y = −0.10 cos (9.9t + .1)

◦
y = 0.10 sin (9.9t + 5)

◦
y = −0.05 cos 9.9t
Physics for Scientists and Engineers ¦ Serway, Jewett ¦ 9th Edition
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Marked as best answer by rosent76 on Aug 6, 2020

kmb352

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Reply #1 on: Aug 6, 2020
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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rosent76

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Reply 2 on: Aug 6, 2020
Great answer, keep it coming :)

Liddy

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Reply 3 on: Yesterday
Thanks for the timely response, appreciate it

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