Question

a small mass m is tied to a ceiling of height h=5l/4 by a string of length L and attached to the floor by a massless spring with constant k=16 mg/L. When the mass is in equilibrium position the spring is stretched by n amount x=L/8. The mass is pulled to the side until the spring has a length s=3L/4 and released from rest. What is the speed of the mass as it passes through its initial position?

Answer 1

Energy of spring in the initial position = \[\frac{ 1 }{ 2}k \left( \frac{ L }{ 8} \right) ^{2}\]
Energy of spring in the stretched position = \[\frac{ 1 }{ 2}k \left( \frac{3 L }{ 4} \right) ^{2}\]
Difference in the PE = KE
\[\frac{ 1 }{ 2 }m v^{2}=\frac{ 1 }{ 2 }k \frac{ 35 }{ 64 }L^{2}\]
\[v^{2}=\frac{ 35kL ^{2} }{ 64m}\]

Answer 2

I appreciate the effort, I'll see how that goes in my group discussions. Thanks.