Object Oscillation & Spring Shortening: Find the Answer!

[leprofece]

1 object of 6.2 Kg hangs 1 balance of spring mass and performs an oscillation of each 0.3 sec If K = 2755 n/m?
How much shortens the spring by removing the object?
Use pi ^ 2 as 10

 

[I like Serena]

1 object of 6.2 Kg hangs 1 balance of spring mass and performs an oscillation of each 0.3 sec If K = 2755 n/m?
How much shortens the spring by removing the object?
Use pi ^ 2 as 10

Can you add some thoughts please?
Perhaps some relevant equations?

 

[leprofece]

Can you add some thoughts please?
Perhaps some relevant equations?

maybe w= 2pi/t
w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I don't get the 0.022 m that is the book answer

it could be get the aceleration to get the force
to try f = Kx

SO CAN anybody help me?

 

[I like Serena]

maybe w= 2pi/t
w= 6,28/0,3

I tried (2pi)sqrt(L/g) solve to L But I don't get the 0.022 m that is the book answer

I'm afraid that is the formula for a pendulum with length L.
The formula for a mass on a spring is
$$\omega = \sqrt{\frac K m}$$

For this problem we won't need it though.
It appears you have more information than you need.

it could be get the aceleration to get the force
to try f = Kx

Good!
Let's pick $y$ for the coordinate though, to emphasize it's a vertical coordinate.
So we have an elastic force $F_e$:
$$F_e = Ky$$
And we also have the force of gravity.
$$F_G = mg$$

When the mass is at rest, the elastic force and the force of gravity have to be equal and opposite.
That is:
$$F_e = F_G$$
$$Ky = mg$$

Now if we remove the mass from the spring, the spring will return to its neutral position at $y=0$.
What can you conclude then about the $y$ where the mass was at rest before?

 

[CellCoree]

.

I would first like to clarify the given information. It seems that the object of 6.2 kg is hanging from a spring with a spring constant (K) of 2755 N/m. The object is performing an oscillation with a period of 0.3 seconds.

To calculate the amount of shortening of the spring, we need to use the equation for the period of oscillation of a spring-mass system: T = 2π√(m/K), where T is the period, m is the mass of the object, and K is the spring constant.

Using the given information, we can rearrange this equation to solve for the spring shortening, which is represented by x:

x = (2π√(m/K) - 2π√(m/K0))^2

Where K0 is the initial value of the spring constant (before the object was added). Using the given values, we can plug them into the equation:

x = (2π√(6.2 kg/2755 N/m) - 2π√(6.2 kg/10))^2

x = (2π√0.0022 - 2π√0.62)^2

x = (0.094 - 3.93)^2 = 3.83^2 = 14.68 cm

Therefore, by removing the object, the spring will shorten by approximately 14.68 cm. This calculation assumes that the spring follows Hooke's Law, which states that the force exerted by the spring is directly proportional to the displacement of the object from its equilibrium position. It is also important to note that this calculation uses an idealized model and may not account for factors such as friction and air resistance, which can affect the actual amount of shortening. Further experimentation and analysis would be needed to determine the exact amount of shortening in a real-world scenario.