The potential energy of a harmonic oscillator of mass 2kg in its equilibrium position is 5 joules. Its total energy is 9 joules and its amplitude is 1cm. Its time period will be

Options

A. 6.28 s
B. `3.14 xx 10^(-2)`s
C. 3.14 s
D. 0.314 s

Answer

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

### Step 1: Understand the given information
- Mass (m) = 2 kg
- Potential Energy (PE) at equilibrium position = 5 Joules
- Total Energy (E) = 9 Joules
- Amplitude (A) = 1 cm = 0.01 m

### Step 2: Calculate the Kinetic Energy (KE) at equilibrium position
At the equilibrium position, the total energy is the sum of potential energy and kinetic energy:
\[ E = PE + KE \]
Substituting the known values:
\[ 9 \, \text{J} = 5 \, \text{J} + KE \]
\[ KE = 9 \, \text{J} - 5 \, \text{J} = 4 \, \text{J} \]

### Step 3: Relate Kinetic Energy to Maximum Velocity
The kinetic energy can be expressed in terms of maximum velocity (V_max):
\[ KE = \frac{1}{2} m V_{\text{max}}^2 \]
Substituting the known values:
\[ 4 \, \text{J} = \frac{1}{2} \times 2 \, \text{kg} \times V_{\text{max}}^2 \]
This simplifies to:
\[ 4 = 1 \times V_{\text{max}}^2 \]
So,
\[ V_{\text{max}}^2 = 4 \implies V_{\text{max}} = \sqrt{4} = 2 \, \text{m/s} \]

### Step 4: Relate Maximum Velocity to Angular Frequency
The maximum velocity in simple harmonic motion is given by:
\[ V_{\text{max}} = \omega A \]
Where \( \omega \) is the angular frequency and \( A \) is the amplitude. Rearranging gives:
\[ \omega = \frac{V_{\text{max}}}{A} \]
Substituting the known values:
\[ \omega = \frac{2 \, \text{m/s}}{0.01 \, \text{m}} = 200 \, \text{rad/s} \]

### Step 5: Calculate the Time Period
The angular frequency is related to the time period (T) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Rearranging gives:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{200} = \frac{\pi}{100} \, \text{s} \]

### Step 6: Numerical Approximation
Using \( \pi \approx 3.14 \):
\[ T \approx \frac{3.14}{100} = 0.0314 \, \text{s} \]

### Final Answer
The time period \( T \) of the harmonic oscillator is approximately \( 0.0314 \, \text{s} \).

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