# Simple Harmonic Motion - Determine the period of oscillation

• KevinFan
In summary, the conversation discusses finding the period of oscillation for a rigid rod suspended from a pivot and set into oscillation. The correct period is 2.19s. For the second part, using omega=sqrt((g/l)), the calculated period is 2.007s, resulting in a percentage difference of 0.0835. However, this answer is incorrect and the correct method is to multiply the difference by 100, which gives a percentage difference of 8.35%. After realizing this, the student thanks for the help.
KevinFan

## Homework Statement

A very light, rigid rod with a length of 0.620m extends straight out from one end of a meter stick. The stick is suspended from a pivot at the far end of the rod and is set into oscillation.
(a)Determine the period of oscillation.
(b)By what percentage does this differ from a one meter long simple pendulum?

## Homework Equations

omega=sqrt((mgh)/i)
omega=sqrt((g/l))
omega=(2Pi)/T

## The Attempt at a Solution

I got the correct period for part a, and it's 2.19s
For the second part, I used omega=sqrt((g/l)), and omega is 3.13, and T is 2.007s.
I used (2.19-2.007)/2.19 to find the percentage which is 0.0835. However, it's wrong...
This seems like a simple question but I'm keep getting the second part wrong. Any suggestions?

I believe the denominator should be the period of the 1 m pendulum .

rock.freak667 said:
I believe the denominator should be the period of the 1 m pendulum .
The system said it's incorrect

KevinFan said:
The system said it's incorrect
Do you mean that you tried that as well, and the system said both answers are incorrect, or are you under the mistaken impression that rock.freak's suggestion is what you originally posted?

haruspex said:
Do you mean that you tried that as well, and the system said both answers are incorrect, or are you under the mistaken impression that rock.freak's suggestion is what you originally posted?
I tried both answers and they are all incorrect

haruspex said:
Do you mean that you tried that as well, and the system said both answers are incorrect, or are you under the mistaken impression that rock.freak's suggestion is what you originally posted?
Never mind, rock.freak667's method is correct. I forgot to multiply my answer by 100...
Thanks for all the help

Good that you found your mistake. Those errors are simple to make and easily overlooked at times!

KevinFan

## 1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where an object moves back and forth in a regular pattern. It is characterized by a restoring force that is proportional to the displacement from an equilibrium position.

## 2. How do you determine the period of oscillation for simple harmonic motion?

The period of oscillation for simple harmonic motion can be determined by dividing the total time for one full cycle by the number of cycles. This can also be calculated using the equation T=2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant.

## 3. What factors affect the period of oscillation in simple harmonic motion?

The period of oscillation is affected by the mass of the object, the spring constant, and the amplitude of the oscillation. A larger mass or stiffer spring will result in a longer period, while a larger amplitude will result in a shorter period.

## 4. How does the period of oscillation change if the amplitude is doubled?

If the amplitude is doubled, the period of oscillation will remain the same. This is because the period is only affected by the mass and spring constant, not the amplitude of the oscillation.

## 5. Can the period of oscillation change during simple harmonic motion?

No, the period of oscillation will remain constant during simple harmonic motion as long as there are no changes to the mass, spring constant, or amplitude of the oscillation. This is because simple harmonic motion is a type of periodic motion, meaning it repeats in a regular pattern.

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