Question

Consider a particle executing a simple harmonic motion. Let x,A,K and U are displacement,amplitude,kinetic energy and potential energy,respectively,of the particle at certain instant of time. If\(\frac{K}{U}\)=3,then \(\frac{x}{A}\) is:

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{9}\)
• E. \(\frac{4}{9}\)

Answer 1

The Correct Option is B

Given:

• Simple harmonic motion with displacement \( x \), amplitude \( A \)
• Ratio of kinetic energy to potential energy: \( \frac{K}{U} = 3 \)

Step 1: Express Energy in SHM

In SHM:

• Total energy \( E = \frac{1}{2}kA^2 \) (constant)
• Potential energy \( U = \frac{1}{2}kx^2 \)
• Kinetic energy \( K = E - U = \frac{1}{2}k(A^2 - x^2) \)

Step 2: Use Given Ratio

Given \( \frac{K}{U} = 3 \):

\[ \frac{\frac{1}{2}k(A^2 - x^2)}{\frac{1}{2}kx^2} = 3 \]

Simplify:

\[ \frac{A^2 - x^2}{x^2} = 3 \]

\[ A^2 - x^2 = 3x^2 \]

\[ A^2 = 4x^2 \]

\[ \frac{x^2}{A^2} = \frac{1}{4} \]

\[ \frac{x}{A} = \frac{1}{2} \]

Conclusion:

The ratio \( \frac{x}{A} \) is \( \frac{1}{2} \).

Answer: \(\boxed{B}\)

Answer 2

1. Potential Energy (U) and Kinetic Energy (K):
  - The potential energy \(U\) at displacement \(x\) is given by:
    \[ U = \frac{1}{2} k x^2   \]
  - The total mechanical energy \(E\) in simple harmonic motion is:
    \[ E = \frac{1}{2} k A^2  \]
  - The kinetic energy \(K\) at displacement \(x\) is given by:
    \[   K = E - U = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2)\]

2. Given Ratio of Kinetic to Potential Energy:
  \[  \frac{K}{U} = 3\]

  Substitute the expressions for \(K\) and \(U\):
  \[\frac{\frac{1}{2} k (A^2 - x^2)}{\frac{1}{2} k x^2} = 3 \]

3. Simplify the Equation:
  \[ \frac{A^2 - x^2}{x^2} = 3\]

4. Solve for \(\frac{x}{A}\):
  - Rearrange the equation:
    \[A^2 - x^2 = 3x^2\]
  - Combine like terms:
    \[A^2 = 4x^2\]
  - Divide both sides by \(A^2\):
    \[1 = 4\left(\frac{x}{A}\right)^2 \]
  - Take the square root of both sides:
    \[ \frac{x}{A} = \pm \frac{1}{2}\]

Thus, the ratio \(\frac{x}{A}\) is \(\pm \frac{1}{2}\).

So, \(\frac{x}{A} = \frac{1}{2}\) or \(\frac{x}{A} = -\frac{1}{2}\).

So the correct Answer is Option B(2):\(\frac12\)

Answer 3

Step 1: Recall the expressions for kinetic and potential energy in simple harmonic motion.

In simple harmonic motion (SHM), the total energy \( E \) is conserved and is given by:

\[ E = K + U, \]

where:

• \( K = \frac{1}{2} m \omega^2 (A^2 - x^2) \) is the kinetic energy,
• \( U = \frac{1}{2} m \omega^2 x^2 \) is the potential energy,
• \( A \) is the amplitude of SHM, and
• \( x \) is the displacement at a given instant.

 

We are given that the ratio of kinetic energy to potential energy is:

\[ \frac{K}{U} = 3. \]

Substitute the expressions for \( K \) and \( U \):

\[ \frac{\frac{1}{2} m \omega^2 (A^2 - x^2)}{\frac{1}{2} m \omega^2 x^2} = 3. \]

Simplify:

\[ \frac{A^2 - x^2}{x^2} = 3. \]

Rearrange:

\[ A^2 - x^2 = 3x^2. \]

Simplify further:

\[ A^2 = 4x^2. \]

Take the square root of both sides:

\[ \frac{x}{A} = \frac{1}{2}. \] ---

Final Answer: The value of \( \frac{x}{A} \) is \( \mathbf{\frac{1}{2}} \), which corresponds to option \( \mathbf{(B)} \).