- #1

songoku

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- Homework Statement
- A mass M is attached to a free pulley and the pulley is attached to wire S and elastic spring (see diagram). The spring will extend 2.0 mm per Newton tensile force. When M is stationary, the extension of spring is 98 mm. The mass of pulley, wire and spring are negligible. M is pushed slowly upwards until there is no spring restoring force, then released.

a. by how far will M move down before it starts going up again? (answer = 392 mm)

b. find the period of the motion (answer = 0.2 π second)

- Relevant Equations
- Hooke's law

Simple harmonic motion

1.

How will the motion of M be? I assume wire S is inelastic so will M move downwards but not in straight line? (I mean M moves downwards but because the left side of pulley is connected to S, it will be static and the right side of pulley can go down along the extension of the spring so its movement will be curve). Or I assume wire S just like spool of thread coiling around the pulley so the pulley and M will move vertically downwards?

2.

Why the answer to question (a) is not 196 mm? When M is stationary and the spring extended by 98 mm, this will be equilibrium point of simple harmonic motion. When M is pushed slowly upwards until there is no spring restoring force, it means that the extension of spring is 0 so I think 98 mm will be the amplitude of simple harmonic motion. When M released, it will move downwards as far as twice the amplitude

Thanks

How will the motion of M be? I assume wire S is inelastic so will M move downwards but not in straight line? (I mean M moves downwards but because the left side of pulley is connected to S, it will be static and the right side of pulley can go down along the extension of the spring so its movement will be curve). Or I assume wire S just like spool of thread coiling around the pulley so the pulley and M will move vertically downwards?

2.

Why the answer to question (a) is not 196 mm? When M is stationary and the spring extended by 98 mm, this will be equilibrium point of simple harmonic motion. When M is pushed slowly upwards until there is no spring restoring force, it means that the extension of spring is 0 so I think 98 mm will be the amplitude of simple harmonic motion. When M released, it will move downwards as far as twice the amplitude

Thanks