# Deriving time peroid of a spring

• thomas49th
In summary, the conversation discussed the derivation of the time period for a spring with a mass m attached to it, which is T = 2pi sqrt(m/k). It was mentioned that the acceleration of a spring is always towards the equilibrium point and the speed can be measured as positive or negative from this point. The spring will travel fastest at the equilibrium point with a displacement of 0. The conversation also included the definition of simple harmonic motion and the equation x(t)= Acos(\sqrt{k/m}t)+ B sin(\sqrt{k/m}t) was shown to satisfy the equation F= ma= mx"= -kx, with periods determined by the constants A and B.
thomas49th
Hi all, I was wondering if we could go through how we get the time peroid for a spring with a mass m attached to it. The time period is:

T = 2pi sqrt(m/k)We know that F = -kx
We know the the the acceleration of a spring is always towards the equilibrium point
the speed can be measured negative and positive from the equilibrium point, as well as the displacement

The spring will be traveling fastest as it goes through the equilibrium point and it's displacement will be 0 at this point

Does that help towards the derivation?

Thanks
Tom

Last edited:
F=ma

and the definition for simple harmonic motion will help

thomas49th said:
Hi all, I was wondering if we could go through how we get the time peroid for a spring with a mass m attached to it. The time period is:

T = 2pi sqrt(m/k)

We know that F = -kx
We know the the the acceleration of a spring is always towards the equilibrium point
the speed can be measured negative and positive from the equilibrium point, as well as the displacement

The spring will be traveling fastest as it goes through the equilibrium point and it's displacement will be 0 at this point

Does that help towards the derivation?

Thanks
Tom
F= ma= mx"= -kx.
Show that, for all numbers A and B,
$$x(t)= Acos(\sqrt{k/m}t)+ B sin(\sqrt{k/m}t)$$
satisfies that equation. What are the periods of those functions?

## 1. How do you calculate the time period of a spring?

The time period of a spring can be calculated using the equation T = 2π√(m/k), where T is the time period, m is the mass attached to the spring, and k is the spring constant.

## 2. What is the significance of the time period of a spring?

The time period of a spring is an important factor in understanding its behavior. It represents the time it takes for the spring to complete one full cycle of oscillation.

## 3. How does the mass attached to a spring affect its time period?

The time period of a spring is directly proportional to the square root of the mass attached to it. This means that as the mass increases, the time period also increases.

## 4. Can the time period of a spring be affected by external factors?

Yes, the time period of a spring can be affected by external factors such as friction, air resistance, and temperature. These factors can alter the oscillation of the spring and therefore affect its time period.

## 5. How can the time period of a spring be measured?

The time period of a spring can be measured by attaching a mass to the spring, pulling it down and releasing it, then using a stopwatch to time the number of oscillations completed in a certain amount of time. The time period can then be calculated by dividing the total time by the number of oscillations.

Replies
6
Views
384
Replies
7
Views
827
Replies
4
Views
1K
Replies
5
Views
1K
Replies
4
Views
1K
Replies
24
Views
2K
Replies
3
Views
765
Replies
17
Views
2K
Replies
2
Views
1K
Replies
3
Views
864