# Pendulum speed with only height from equilibrium

• Serpentia
In summary, the problem involves a pendulum being dropped from a height of 0.50 metres above its equilibrium position. The length of the string is 1.0m and the acceleration due to gravity is 9.81 m/s^2. Using the formula for mechanical energy, the potential energy at 0.05 metres is equal to the kinetic energy at the equilibrium position. By setting these two energies equal to each other and solving for velocity, the speed of the pendulum as it passes through the equilibrium position can be determined.
Serpentia

## Homework Statement

A pendulum is dropped from 0.50 metres above the equilibrium position. What is the speed of the pendulum as it passes through the equilibrium position? (The pendulum is in a isolated system.)

h1= 0.05m
h2= 0.00m
g= 9.81 m/s^2
Length of string (L)= 1.0m.

## Homework Equations

Em=Ek+Ep
Em=[(1/2)mv^2]+[mgh]
v=√(2Ek/m)

## The Attempt at a Solution

I'm not entirely sure what to try. I'm not given mass, or any value of any form of Energy, be it Kinetic or Gravitational Potential. I've tried to find the value of mass, but I don't have the energy or velocity factor to figure it out;

v=√(2Ek/m)
m=(2Ek/v^2), or
m=(Ep/gh)

Just don't work because I don't have that information. I'm totally lost... And that's all the formulas I can extrapolate from what information they've given me. All I need is a formula to follow, and I can get it from there, thanks muchly.

At 0.05 metres what is the potential energy?

This same energy is converted entirely to ke as it passes through the equilibrium position.

1/2 mv2=mgh

rock.freak667 said:
At 0.05 metres what is the potential energy?

This same energy is converted entirely to ke as it passes through the equilibrium position.

1/2 mv2=mgh

I'm not supplied potential, kinetic, or mechanical energy.

Serpentia said:
I'm not supplied potential, kinetic, or mechanical energy.

Using the formula, m cancels out. you have g and h. find v.

rock.freak667 said:
Using the formula, m cancels out. you have g and h. find v.

Oh, doy, thanks so much. Got it now.

## 1. What is a pendulum and how does it work?

A pendulum is a weight suspended from a fixed point so that it can swing freely back and forth. The motion of a pendulum is governed by the force of gravity and the length of the pendulum. As the pendulum swings, it transfers energy between potential energy (when it is at its highest point) and kinetic energy (when it is at its lowest point).

## 2. How does the height from equilibrium affect the speed of a pendulum?

The height from equilibrium, also known as the amplitude, directly affects the speed of a pendulum. The higher the amplitude, the faster the pendulum will swing. This is because a higher amplitude means the pendulum has more potential energy at its highest point and therefore, more kinetic energy at its lowest point.

## 3. Can the speed of a pendulum be calculated based on its height from equilibrium?

Yes, the speed of a pendulum can be calculated using the equation v = √(2gh), where v is the speed, g is the acceleration due to gravity (9.8 m/s²), and h is the height from equilibrium. This equation assumes that there is no friction or air resistance, and the pendulum is swinging in a vacuum.

## 4. Does the length of a pendulum affect its speed with only height from equilibrium?

No, the length of a pendulum does not affect its speed with only height from equilibrium. As long as the amplitude remains the same, the speed of a pendulum will not change regardless of its length. However, the length of a pendulum does affect its period (the time it takes to complete one swing) and frequency (the number of swings per unit of time).

## 5. How can the speed of a pendulum with only height from equilibrium be used in real-world applications?

The speed of a pendulum with only height from equilibrium has various real-world applications, such as determining the time in pendulum clocks, measuring gravity, and studying the principles of energy conservation. It is also used in seismology to measure the strength of earthquakes and in sports for training and timing purposes.

• Introductory Physics Homework Help
Replies
9
Views
680
• Introductory Physics Homework Help
Replies
27
Views
710
• Introductory Physics Homework Help
Replies
9
Views
389
• Introductory Physics Homework Help
Replies
26
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
856
• Introductory Physics Homework Help
Replies
9
Views
2K
• Classical Physics
Replies
18
Views
839
• Introductory Physics Homework Help
Replies
2
Views
713
• Introductory Physics Homework Help
Replies
7
Views
2K
• Mechanical Engineering
Replies
19
Views
1K