Question

When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is \(\frac{x}{ 8}\) , where x = _________.

Answer 1

Correct Answer: 9

Step 1: Define Total Energy - The total energy E of a simple harmonic oscillator is given by:

E = \(\frac{1}{2}KA^2\)

- Where K is the spring constant and A is the amplitude.

Step 2: Calculate Potential Energy U at Displacement \(\frac{A}{3}\) - When the displacement is \(\frac{A}{3}\):

U = \(\frac{1}{2}K \left(\frac{A}{3}\right)^2 = \frac{KA^2}{18} = \frac{E}{9}\)

Step 3: Calculate Kinetic Energy KE - Kinetic energy is the difference between total energy and potential energy:

KE = E - U = E - \(\frac{E}{9} = \frac{8E}{9}\)

Step 4: Calculate the Ratio of Total Energy to Kinetic Energy:

\(\frac{\text{Total Energy}}{\text{KE}} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}\)

Step 5: Determine x - Since the ratio is \(\frac{x}{8}\), we have x = 9.

So, the correct answer is: x = 9

Answer 2

Step 1: Total energy in SHM.

\[ E = \frac{1}{2} k A^2 \] where \( A \) is the amplitude.

Step 2: Kinetic energy at displacement \( x \).

In SHM, potential energy at displacement \( y \) is: \[ U = \frac{1}{2} k y^2 \] Hence, kinetic energy: \[ K = E - U = \frac{1}{2}kA^2 - \frac{1}{2}ky^2 = \frac{1}{2}k(A^2 - y^2) \]

Step 3: Substitute \( y = \frac{A}{3} \).

\[ K = \frac{1}{2}k\left(A^2 - \frac{A^2}{9}\right) \] \[ K = \frac{1}{2}kA^2\left(\frac{8}{9}\right) \]

Step 4: Ratio of total energy to kinetic energy.

\[ \frac{E}{K} = \frac{\frac{1}{2}kA^2}{\frac{1}{2}kA^2 \times \frac{8}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8} \]

Final Answer:

\[ \boxed{x = 9} \]