The potential energy of a harmonic oscillator of mass 2 kg in its mean positioin is 5J. If its total energy is 9J and its amplitude is 0.01m, its period will be

Options

A. `(pi/100)s`
B. `(pi)/(50)s`
C. `(pi)/(20)s`
D. `(pi)/10)`s

Answer

To find the period of the harmonic oscillator, we will follow these steps:

### Step 1: Understand the relationship between total energy, potential energy, and kinetic energy.
The total mechanical energy (E) of a harmonic oscillator is given by the sum of its potential energy (PE) and kinetic energy (KE):
\[ E = PE + KE \]

### Step 2: Calculate the kinetic energy at the mean position.
Given:
- Total energy \( E = 9 \, J \)
- Potential energy at mean position \( PE = 5 \, J \)

Using the formula:
\[ KE = E - PE \]
Substituting the values:
\[ KE = 9 \, J - 5 \, J = 4 \, J \]

### Step 3: Relate kinetic energy to maximum velocity.
The kinetic energy can also be expressed in terms of mass and maximum velocity:
\[ KE = \frac{1}{2} m v^2 \]
Where:
- \( m = 2 \, kg \) (mass of the oscillator)
- \( v \) is the maximum velocity.

Setting the equations equal:
\[ 4 \, J = \frac{1}{2} \times 2 \, kg \times v^2 \]
This simplifies to:
\[ 4 = v^2 \]
Thus:
\[ v = \sqrt{4} = 2 \, m/s \]

### Step 4: Relate maximum velocity to angular frequency and amplitude.
The maximum velocity \( v \) is also related to angular frequency \( \omega \) and amplitude \( A \):
\[ v = \omega A \]
Given that the amplitude \( A = 0.01 \, m \), we can write:
\[ 2 = \omega \times 0.01 \]

### Step 5: Solve for angular frequency \( \omega \).
Rearranging the equation gives:
\[ \omega = \frac{2}{0.01} = 200 \, rad/s \]

### Step 6: Find the period \( T \) using angular frequency.
The relationship between angular frequency and period is:
\[ \omega = \frac{2\pi}{T} \]
Rearranging this gives:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{200} = \frac{\pi}{100} \, s \]

### Final Answer:
The period \( T \) of the harmonic oscillator is:
\[ T = \frac{\pi}{100} \, s \]

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