Two exactly identical simple pendulums are oscillating with amplitude 2 cma dn 6cm Calculate the ratio of their energies of oscillation.

Answer

To find the ratio of the energies of two identical simple pendulums oscillating with different amplitudes, we can follow these steps:

### Step 1: Understand the formula for energy in simple harmonic motion
The energy \( E \) of a simple harmonic oscillator (like a pendulum) is given by the formula:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( E \) is the energy,
- \( m \) is the mass of the pendulum,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of oscillation.

### Step 2: Identify the amplitudes of the two pendulums
From the problem statement, we have:
- Amplitude of the first pendulum \( A_1 = 2 \, \text{cm} \)
- Amplitude of the second pendulum \( A_2 = 6 \, \text{cm} \)

### Step 3: Write the expressions for the energies of both pendulums
Using the energy formula:
- For the first pendulum:
\[ E_1 = \frac{1}{2} m \omega^2 A_1^2 = \frac{1}{2} m \omega^2 (2 \, \text{cm})^2 \]
- For the second pendulum:
\[ E_2 = \frac{1}{2} m \omega^2 A_2^2 = \frac{1}{2} m \omega^2 (6 \, \text{cm})^2 \]

### Step 4: Calculate the ratio of the energies
Now, we can find the ratio of the energies \( \frac{E_1}{E_2} \):
\[ \frac{E_1}{E_2} = \frac{\frac{1}{2} m \omega^2 (2 \, \text{cm})^2}{\frac{1}{2} m \omega^2 (6 \, \text{cm})^2} \]
The \( \frac{1}{2} m \omega^2 \) terms cancel out:
\[ \frac{E_1}{E_2} = \frac{(2 \, \text{cm})^2}{(6 \, \text{cm})^2} = \frac{4 \, \text{cm}^2}{36 \, \text{cm}^2} = \frac{1}{9} \]

### Step 5: Conclusion
Thus, the ratio of the energies of the two pendulums is:
\[ \frac{E_1}{E_2} = \frac{1}{9} \]