The ratio of maximum velocity to the velocity of a particle performing S.H M at a point where potential energy is 25% of total energy is

Options

A. `2:sqrt3`
B. `sqrt3 : 2`
C. `1 : 2`
D. `1 : sqrt3`

Answer

To solve the problem of finding the ratio of maximum velocity to the velocity of a particle performing simple harmonic motion (S.H.M) at a point where the potential energy is 25% of the total energy, we can follow these steps:

### Step 1: Understand the relationship between potential energy and total energy in S.H.M.
In simple harmonic motion, the total mechanical energy (E) is the sum of kinetic energy (K.E) and potential energy (P.E). The total energy can be expressed as:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Express the potential energy when it is 25% of total energy.
According to the problem, the potential energy at a certain point is 25% of the total energy:
\[ P.E = 0.25 E = \frac{1}{4} \left( \frac{1}{2} m \omega^2 A^2 \right) = \frac{1}{8} m \omega^2 A^2 \]

### Step 3: Relate potential energy to displacement.
The potential energy in S.H.M can also be expressed as:
\[ P.E = \frac{1}{2} m \omega^2 x^2 \]
Setting the two expressions for potential energy equal gives:
\[ \frac{1}{2} m \omega^2 x^2 = \frac{1}{8} m \omega^2 A^2 \]

### Step 4: Simplify the equation.
We can cancel \( m \omega^2 \) from both sides (assuming \( m \) and \( \omega \) are not zero):
\[ \frac{1}{2} x^2 = \frac{1}{8} A^2 \]
Multiplying both sides by 8 gives:
\[ 4x^2 = A^2 \]
Thus,
\[ x^2 = \frac{A^2}{4} \]
Taking the square root:
\[ x = \frac{A}{2} \]

### Step 5: Find the maximum velocity.
The maximum velocity \( V_{max} \) occurs at the mean position (x = 0):
\[ V_{max} = A \omega \]

### Step 6: Find the velocity at \( x = \frac{A}{2} \).
The velocity \( v \) at displacement \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
Substituting \( x = \frac{A}{2} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} \]
\[ v = \omega \sqrt{A^2 - \frac{A^2}{4}} \]
\[ v = \omega \sqrt{\frac{3A^2}{4}} \]
\[ v = \frac{\omega A \sqrt{3}}{2} \]

### Step 7: Calculate the ratio of maximum velocity to the velocity at \( x = \frac{A}{2} \).
Now we can find the ratio:
\[ \text{Ratio} = \frac{V_{max}}{v} = \frac{A \omega}{\frac{\omega A \sqrt{3}}{2}} \]
Canceling \( A \omega \):
\[ \text{Ratio} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]

### Final Answer:
The ratio of maximum velocity to the velocity of the particle at the point where potential energy is 25% of total energy is:
\[ \frac{2}{\sqrt{3}} \]

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