# the potential energy of an occilator of mass 2 kg in its mean position is 5 joules...if its total energy is 9 joules and apmlitute of 0.01 m....its tym period is?????

Swapnil Saxena
102 Points
12 years ago

The Potential Energy of the paricle is 5 Joules at the mean pos.(With Reference to an arbitary Refrence Point).

With Reference to same arbitary Refrence Point , the total energy of the particle is 9 Joules.

So the Kinetic Energies at the mean pos. = 9J -5J = 4 Joules

At the amplitude the all the kinetic energy will convert into spring potential energy

4 Joules = (1/2)kx2 = (1/2)*k(0.01)2

k = 8*10

Time Period is 2pi (m/k)1/2= 2pi (2/8*104)1/2 = 2pi(0.25*10-4)1/2= 2 pi (0.5*10-2)=3.14*10-2s
The text passage contains a physics problem related to the potential and kinetic energy of an oscillator, and it provides a detailed solution to find the time period of the oscillator.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# the potential energy of an occilator of mass 2 kg in its mean position is 5 joules...if its total energy is 9 joules and apmlitute of 0.01 m....its tym period is?????

Swapnil Saxena
102 Points
12 years ago

The Potential Energy of the paricle is 5 Joules at the mean pos.(With Reference to an arbitary Refrence Point).

With Reference to same arbitary Refrence Point , the total energy of the particle is 9 Joules.

So the Kinetic Energies at the mean pos. = 9J -5J = 4 Joules

At the amplitude the all the kinetic energy will convert into spring potential energy

4 Joules = (1/2)kx2 = (1/2)*k(0.01)2

k = 8*10

Time Period is 2pi (m/k)1/2= 2pi (2/8*104)1/2 = 2pi(0.25*10-4)1/2= 2 pi (0.5*10-2)=3.14*10-2s
user
The text passage contains a physics problem related to the potential and kinetic energy of an oscillator, and it provides a detailed solution to find the time period of the oscillator.

TYPE: SINGLE_QUESTION_WITH_ANSWER