Vertical spring-mass system

  • #1
Ark236
22
3
Homework Statement
Hi everyone,

The problem has two parts. The first is to determine the equilibrium position of a mass attached to a spring. The second is to determine the equation of motion of the system, assuming that the block is pulled 1 cm down from its equilibrium position.
Relevant Equations
I choose the downward direction as positive. For the first part and using the FBD:

mg - k y_{0} = 0

Then the equilibrium position is y_{0} = mg/k.

For the second part, we have that:

mg -k y = m d^2 y/dt^2
I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.

thanks

image was obtained from here

Thanks.
C_3oscilador.png
 
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  • #2
Ark236 said:
Homework Statement: Hi everyone,

The problem has two parts. The first is to determine the equilibrium position of a mass attached to a spring. The second is to determine the equation of motion of the system, assuming that the block is pulled 1 cm down from its equilibrium position.
Relevant Equations: I choose the downward direction as positive. For the first part and using the FBD:

mg - k y_{0} = 0

Then the equilibrium position is y_{0} = mg/k.

For the second part, we have that:

mg -k y = m d^2 y/dt^2

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.

thanks

image was obtained from here

Thanks.View attachment 334759
When ##y>y_0, mg-ky<0##, and then the right side, ##m \frac {d^2y} {dt^2}## is negative.
 
  • #3
Both sides are negative at that point. The net force points upwards and the body accelerates upwards. The equation shows that thenet force and the acceleration have the same sign. They are either both positive or both negative. If you put a minus sign in the equation itself, it would mean that the acceleration is in direction opposite to the net force. This would contradict Newton's second law, wouldn't?
 
  • #4
Ark236 said:
mg -k y = m d^2 y/dt^2

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards.
When the mass is below it's equilibrium position, which is bigger: mg or ky?
 

1. What is a vertical spring-mass system?

A vertical spring-mass system is a physical system consisting of a mass attached to a spring that is fixed at one end and hanging vertically from the other end. The mass undergoes simple harmonic motion as it oscillates up and down due to the force of gravity and the restoring force of the spring.

2. What factors affect the motion of a vertical spring-mass system?

The motion of a vertical spring-mass system is affected by the mass of the object, the spring constant of the spring, and the amplitude of the oscillation. The force of gravity and any external forces acting on the system can also affect its motion.

3. How is the period of oscillation calculated for a vertical spring-mass system?

The period of oscillation for a vertical spring-mass system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. This equation assumes that there is no external damping or friction.

4. How does the spring constant affect the frequency of a vertical spring-mass system?

The spring constant directly affects the frequency of a vertical spring-mass system. A higher spring constant will result in a higher frequency, meaning the mass will oscillate more times per unit of time. Similarly, a lower spring constant will result in a lower frequency.

5. What is the relationship between the amplitude and energy of a vertical spring-mass system?

The amplitude and energy of a vertical spring-mass system are directly proportional. As the amplitude of the oscillation increases, so does the energy of the system. This is because the potential energy stored in the spring increases as the amplitude increases, and this energy is converted to kinetic energy as the mass oscillates.

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