A certain simple harmonic vibrator of mass 0.1 kg has a total energy of 10 J. Its displacement from the mean position is 1 cm when it has equal kinetic and potential energies. The amplitude A and frequency n of vibration of the vibrator are

Options

A. `A=sqrt2cm`,`n=(500)/(pi)Hz`
B. `A=sqrt2cm`,`n=(1000)/(pi)Hz`
C. `A=(1)/(sqrt2)cm`,`n=(500)/(pi)Hz`
D. `A=(1)/(sqrt2)cm`,`n=(1000)/(pi)`Hz

Explanation

To solve the problem step by step, we will find the amplitude \( A \) and frequency \( n \) of the simple harmonic vibrator.

### Given:
- Mass \( m = 0.1 \, \text{kg} \)
- Total energy \( E = 10 \, \text{J} \)
- Displacement \( x = 1 \, \text{cm} = 0.01 \, \text{m} \)
- At this displacement, kinetic energy \( K \) equals potential energy \( U \).

### Step 1: Relate Kinetic and Potential Energy
Since the kinetic energy equals potential energy at \( x = 0.01 \, \text{m} \), we have:
\[ K = U \]

The total energy \( E \) can be expressed as:
\[ E = K + U = 2K \]
Thus, we can write:
\[ K = \frac{E}{2} = \frac{10}{2} = 5 \, \text{J} \]

### Step 2: Write the Expressions for Kinetic and Potential Energy
The kinetic energy \( K \) is given by:
\[ K = \frac{1}{2} m v^2 \]
The potential energy \( U \) is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

### Step 3: Relate \( k \) to \( \omega \)
In simple harmonic motion, the spring constant \( k \) can be related to angular frequency \( \omega \) as:
\[ k = m \omega^2 \]
Substituting this into the potential energy equation gives:
\[ U = \frac{1}{2} m \omega^2 x^2 \]

### Step 4: Set Up the Equation
Since \( K = U \), we can write:
\[ \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 x^2 \]
At maximum displacement \( A \), the total energy is:
\[ E = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \]
Thus, we can express the total energy as:
\[ 10 = \frac{1}{2} m \omega^2 A^2 \]

### Step 5: Find the Amplitude \( A \)
From the equality of kinetic and potential energy at \( x = 0.01 \, \text{m} \):
\[ 5 = \frac{1}{2} m \omega^2 (0.01)^2 \]
Substituting \( m = 0.1 \):
\[ 5 = \frac{1}{2} (0.1) \omega^2 (0.01)^2 \]
\[ 5 = 0.0005 \omega^2 \]
\[ \omega^2 = \frac{5}{0.0005} = 10000 \]
\[ \omega = 100 \, \text{rad/s} \]

Now substituting \( \omega \) back to find \( A \):
\[ 10 = \frac{1}{2} (0.1) (100)^2 A^2 \]
\[ 10 = 0.5 \cdot 10000 \cdot A^2 \]
\[ 10 = 5000 A^2 \]
\[ A^2 = \frac{10}{5000} = 0.002 \]
\[ A = \sqrt{0.002} = 0.04472 \, \text{m} \approx 4.47 \, \text{cm} \]

### Step 6: Find the Frequency \( n \)
The frequency \( n \) can be calculated as:
\[ n = \frac{\omega}{2\pi} \]
Substituting \( \omega = 100 \):
\[ n = \frac{100}{2\pi} \approx \frac{100}{6.2832} \approx 15.92 \, \text{Hz} \]

### Final Answers:
- Amplitude \( A \approx 4.47 \, \text{cm} \)
- Frequency \( n \approx 15.92 \, \text{Hz} \)